23.20 - Solved Examples on Standard Integrals

In the previous section, we saw the derivation of three standard integrals. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved Example 23.63
Find $\small{\int{\left[\sqrt{4 - x^2} \right]dx}}$
Solution:
1. The given integral can be written as: $\small{\int{\left[\sqrt{2^2 - x^2} \right]dx}}$

2. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

Here a = 2

• So we get:

$\small{\int{\left[\sqrt{2^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{2^2 - x^2} ~+~\frac{2^2}{2} \sin^{-1}\frac{x}{2}~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{4 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{4 - x^2} ~+~2 \sin^{-1}\frac{x}{2}~+~\rm{C}}$

Solved Example 23.64
Find $\small{\int{\left[\sqrt{1 - 4 x^2} \right]dx}}$
Solution:
1. The given integral can be written as:

$\small{\int{\left[\sqrt{1^2 - (2x)^2} \right]dx}}$

2. Put u = 2x Then du/dx = 2 ⇒ du = 2 dx

• So the given integral becomes:

$\small{\int{\left[\sqrt{1^2 - (2x)^2} \right]dx}~=~\int{\left[\frac{2 \sqrt{1^2 - (2x)^2}}{2} \right]dx}}$

$\small{~=~\int{\left[\frac{ \sqrt{1^2 - u^2}}{2} \right]du}~=~\frac{1}{2} \int{\left[\sqrt{1^2 - u^2} \right]dx}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

Here a = 1

• So we get:

$\small{\frac{1}{2}\int{\left[\sqrt{1^2 - u^2} \right]du}~=~~\frac{u}{4}\sqrt{1^2 - u^2} ~+~\frac{1^2}{4} \sin^{-1}\frac{u}{1}~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\frac{1}{2}\int{\left[\sqrt{1^2 - (2x)^2} \right]dx}~=~~\frac{2x}{4}\sqrt{1^2 - (2x)^2} ~+~\frac{1}{4} \sin^{-1}(2x)~+~\rm{C}}$

$\small{\Rightarrow \frac{1}{2}\int{\left[\sqrt{1 - 4x^2} \right]dx}~=~~\frac{x}{2}\sqrt{1 - 4x^2} ~+~\frac{1}{4} \sin^{-1}(2x)~+~\rm{C}}$

Solved Example 23.65
Find $\small{\int{\left[\sqrt{x^2 + 4x + 1} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +4x+ 4 - 3} \right]dx}~=~\int{\left[\sqrt{(x+2)^2 - (\sqrt{3})^2} \right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2 - (\sqrt{3})^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

Here x = u and a = √3

• So we get:

$\small{\int{\left[\sqrt{u^2 - (\sqrt{3})^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 - (\sqrt{3})^2} ~-~\frac{(\sqrt{3})^2}{2} \log \left|u+\sqrt{u^2 - (\sqrt{3})^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+2)^2 - (\sqrt{3})^2} \right]dx}~=~\frac{x+2}{2}\sqrt{(x+2)^2 - (\sqrt{3})^2} ~-~\frac{(\sqrt{3})^2}{2} \log \left|(x+2)-\sqrt{(x+2)^2 + (\sqrt{3})^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x + 1} \right]dx}~=~\frac{x+2}{2}\sqrt{x^2 + 4x + 1} ~-~ \frac{3}{2} \log \left|x+2-\sqrt{x^2 + 4x + 1} \right|~+~\rm{C}}$

Solved Example 23.66
Find $\small{\int{\left[\sqrt{x^2 + 4x - 5} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +4x+ 4 - 9} \right]dx}~=~\int{\left[\sqrt{(x+2)^2 - 3^2} \right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2 - 3^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

Here x = u and a = 3

• So we get:

$\small{\int{\left[\sqrt{u^2 - 3^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 - 3^2} ~-~\frac{3^2}{2} \log \left|u+\sqrt{u^2 - 3^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+2)^2 - 3^2} \right]dx}}$

$\small{~=~\frac{x+2}{2}\sqrt{(x+2)^2 - 3^2} ~-~\frac{3^2}{2} \log \left|(x+2)-\sqrt{(x+2)^2 + 3^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x - 5} \right]dx}}$

$\small{~=~\frac{x+2}{2}\sqrt{x^2 + 4x - 5} ~-~ \frac{9}{2} \log \left|x+2-\sqrt{x^2 + 4x - 5} \right|~+~\rm{C}}$

Solved Example 23.67
Find $\small{\int{\left[\sqrt{3 - 2x - x^2}\right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{-(x^2 +2x- 3)} \right]dx}~=~\int{\left[\sqrt{-[x^2 +2x+1-4]}\right]dx}}$

$\small{~=~\int{\left[\sqrt{-\left[(x+1)^2-2^2\right]}\right]dx}~=~\int{\left[\sqrt{2^2 - (x+1)^2}\right]dx}}$

2. Put u = x+1 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{2^2 - u^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

Here a = 2 and x = u

• So we get:

$\small{\int{\left[\sqrt{2^2 - u^2} \right]dx}~=~~\frac{u}{2}\sqrt{2^2 - u^2} ~+~\frac{2^2}{2} \sin^{-1}\frac{u}{2}~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{2^2 - (x+1)^2} \right]dx}}$

$\small{~=~~\frac{x+1}{2}\sqrt{2^2 - (x+1)^2} ~+~\frac{2^2}{2} \sin^{-1}\frac{x+1}{2}~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{3-2x-x^2} \right]dx}}$

$\small{~=~\frac{x+1}{2}\sqrt{3-2x-x^2}~+~2 \sin^{-1}\frac{x+1}{2}~+~\rm{C}}$

Solved Example 23.68
Find $\small{\int{\left[\sqrt{1 - 4x - x^2}\right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{-(x^2 +4x- 1)} \right]dx}~=~\int{\left[\sqrt{-[x^2 +4x+4-5]}\right]dx}}$

$\small{~=~\int{\left[\sqrt{-\left[(x+2)^2-(\sqrt{5})^2\right]}\right]dx}~=~\int{\left[\sqrt{(\sqrt{5})^2 - (x+2)^2}\right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{(\sqrt{5})^2 - u^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

Here a = √5 and x = u

• So we get:

$\small{\int{\left[\sqrt{(\sqrt{5})^2 - u^2} \right]dx}~=~~\frac{u}{2}\sqrt{(\sqrt{5})^2 - u^2} ~+~\frac{(\sqrt{5})^2}{2} \sin^{-1}\frac{u}{\sqrt{5}}~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(\sqrt{5})^2 - (x+2)^2} \right]dx}}$

$\small{~=~~\frac{x+2}{2}\sqrt{(\sqrt{5})^2 - (x+2)^2} ~+~\frac{(\sqrt{5})^2}{2} \sin^{-1}\frac{x+2}{\sqrt{5}}~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{1-4x-x^2} \right]dx}}$

$\small{~=~\frac{x+2}{2}\sqrt{1-4x-x^2}~+~\frac{5}{2} \sin^{-1}\frac{x+2}{\sqrt{5}}~+~\rm{C}}$

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The link below gives a few more solved examples:

Exercise 23.7

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We have completed a discussion on the three standard integrals. In the next section, we will see definite integrals.

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23.19 - Derivation of Some Standard Integrals

In the previous section, we completed a discussion on the method of integration by parts. In this section, we will see three standard integrals.

I. The integral $\bf{\int{\left[\sqrt{x^2 - a^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{x^2 - a^2}}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 - a^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2-a^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2-a^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2-a^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{x^2}{\sqrt{x^2-a^2}}  \big]dx}~=~\int{\big[\frac{x^2 - a^2 + a^2}{\sqrt{x^2-a^2}}  \big]dx}~=~\int{\big[\frac{x^2 - a^2}{\sqrt{x^2-a^2}}~+~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2-a^2}~+~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{x^2 - a^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \sqrt{x^2 - a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} \bigg]}$

$\small{~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\sqrt{x^2-a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{x^2-a^2} \big]dx}~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2-a^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula IV that we saw in section 23.8. Thus we get:

$\small{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

II. The integral $\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{x^2 + a^2}}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 + a^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2+a^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2+a^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2+a^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{x^2}{\sqrt{x^2+a^2}}  \big]dx}~=~\int{\big[\frac{x^2 + a^2 - a^2}{\sqrt{x^2+a^2}}  \big]dx}~=~\int{\big[\frac{x^2 + a^2}{\sqrt{x^2-a^2}}~-~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2+a^2}~-~\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \sqrt{x^2 + a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} \bigg]}$

$\small{~=~x \sqrt{x^2 + a^2} ~-~\int{\big[\sqrt{x^2+a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{x^2+a^2} \big]dx}~=~x \sqrt{x^2 + a^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2+a^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula VI that we saw in section 23.8. Thus we get:

$\small{ \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

III. The integral $\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{a^2 - x^2}}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{a^2 - x^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-x}{\sqrt{a^2-x^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{-x}{\sqrt{a^2-x^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}}  \big]dx}~=~\int{\big[\frac{a^2 - x^2 - a^2}{\sqrt{a^2-x^2}}  \big]dx}~=~\int{\big[\frac{a^2 - x^2}{\sqrt{a^2-x^2}}~-~\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{a^2-x^2}~-~\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x\sqrt{a^2 - x^2} \big]~-~\bigg[\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx} \bigg]}$

$\small{~=~x\sqrt{a^2 - x^2} ~-~\int{\big[\sqrt{a^2-x^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{a^2-x^2} \big]dx}~=~x\sqrt{a^2 - x^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{a^2-x^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula V that we saw in section 23.8. Thus we get:

$\small{ \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

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Now we will see some solved examples.

Solved example 23.61
Find $\small{\int{\left[\sqrt{x^2 +2x+ 5} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +2x+ 1+ 4} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ 4} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ 2^2} \right]dx}}$

2. Put u = x+1 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ 4} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

Here x = u and a = 2

• So we get:

$\small{\int{\left[\sqrt{u^2 + 2^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 + 2^2} ~+~\frac{2^2}{2} \log \left|u+\sqrt{u^2 + 2^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+1)^2 + 2^2} \right]dx}~=~\frac{x+1}{2}\sqrt{(x+1)^2 + 2^2} ~+~\frac{2^2}{2} \log \left|(x+1)+\sqrt{(x+1)^2 + 2^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 2x + 5} \right]dx}~=~\frac{x+1}{2}\sqrt{x^2 + 2x + 5} ~+~2 \log \left|(x+1)+\sqrt{x^2 + 2x + 5} \right|~+~\rm{C}}$

Solved Example 23.62
Find $\small{\int{\left[\sqrt{x^2 +4x+ 6} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +4x+ 4+ 2} \right]dx}~=~\int{\left[\sqrt{(x+2)^2+ 2} \right]dx}~=~\int{\left[\sqrt{(x+1)^2+ (\sqrt{2})^2} \right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ (\sqrt{2})^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

Here x = u and a = √2

• So we get:

$\small{\int{\left[\sqrt{u^2 + (\sqrt{2})^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 + (\sqrt{2})^2} ~+~\frac{(\sqrt{2})^2}{2} \log \left|u+\sqrt{u^2 + (\sqrt{2})^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+2)^2 + (\sqrt{2})^2} \right]dx}~=~\frac{x+2}{2}\sqrt{(x+1)^2 + (\sqrt{2})^2} ~+~\frac{(\sqrt{2})^2}{2} \log \left|(x+2)+\sqrt{(x+2)^2 + (\sqrt{2})^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x + 6} \right]dx}~=~\frac{x+2}{2}\sqrt{x^2 + 4x + 6} ~+~ \log \left|x+2+\sqrt{x^2 + 4x + 6} \right|~+~\rm{C}}$

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We have seen the three standard integrals. In the next section, we will see a few more solved examples.

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23.18 - Integration by Parts - Special case

In the previous section, we saw the method of Integration by parts. We saw some solved examples also. In this section, we will see a special case of this method.

Integral of the type $\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

This can be calculated in 3 steps:

1. We have: $\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

$\small{~=~\int{\left[e^x\,f(x)\right]dx}~+~\int{\left[e^x\,f'(x)\right]dx}}$

• There are two terms. We will denote the first one as I₁.

2. Let us calculate I₁:

(i) Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{f(x)}$

   ♦ Let second function be: g(x) = $\small{e^x}$

(ii) Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^x \right]dx}~=~e^x}$

(iii) $\small{\big[f(x) \left(A \right) \big]~=~\big[f(x) \, \left(e^x \right) \big]}$

• This is the first term.

(iv) $\small{f'(x)~=~f'(x)}$

(v) $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[f'(x)\,\left(e^x \right)  \big]dx}}$

• This is second term.

(vi) So we get:

I₁ = $\small{\text{First term - Second term}}$

$\small{~=~\big[f(x) \, \left(e^x \right) \big]~-~\bigg[\int{\big[f'(x)\,\left(e^x \right)  \big]dx} \bigg]}$

3. So from (1), we get:

$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

$\small{=~\int{\left[e^x\,f(x)\right]dx}~+~\int{\left[e^x\,f'(x)\right]dx}}$

= I₁ + $\small{\int{\left[e^x\,f'(x)\right]dx}}$

= $\big[f(x) \, \left(e^x \right) \big]~-~\bigg[\int{\big[f'(x)\,\left(e^x \right)  \big]dx} \bigg]$ + $\small{\int{\left[e^x\,f'(x)\right]dx}}$

= $\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

---

Let us see some solved examples:

Solved Example 23.57
Find $\small{\int{\big[e^x \left[\sin x + \cos x \right]\big]dx}}$
Solution:
1. Let f(x) = $\small{\sin x}$

• Then $\small{f'(x)~=~\cos x}$

2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \sin x~+~\rm{C}}$

Solved Example 23.58
Find $\small{\int{\left[e^x\,\sec x (1 + \tan x) \right]dx}}$
Solution:
1. The given integral can be written as:

$\small{\int{\big[e^x \left[\sec x + \sec x \tan x \right]\big]dx}}$

Let f(x) = $\small{\sec x}$

• Then $\small{f'(x)~=~\sec x \tan x}$

2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \sec x~+~\rm{C}}$

Solved Example 23.59
Find $\small{\int{\left[\frac{x e^x}{(1+x)^2} \right]dx}}$
Solution:
1. The $\small{\left[\frac{x}{(1+x)^2} \right]}$ portion can be rearranged as shown below:

$\small{\frac{x}{(1+x)^2} = \frac{1+x-1}{(1+x)^2} = \big[\frac{1+x}{(1+x)^2}~+~\frac{(-1)}{(1+x)^2}\big] = \big[\frac{1}{1+x}~+~\frac{(-1)}{(1+x)^2}\big]}$

2. Now, $\small{\frac{(-1)}{(1+x)^2}}$ is the derivative of $\small{\frac{1}{1+x}}$.

(The reader may verify this by doing the differentiation)

3. So the given integral is of the form:

$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

Where $\small{f(x) = \frac{1}{1+x}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \left[\frac{1}{1+x} \right]~+~\rm{C}}$

Solved example 23.60
Find $\small{(i)~\int{\bigg[e^x \left[\tan^{-1}x \,+\,\frac{1}{1+x^2} \right]\bigg]dx}~~~~~~(ii)~\int{\left[\frac{(x^2 + 1)e^x}{(x+1)^2} \right]dx}}$
Solution:
Part (i):
1. Let f(x) = $\small{\tan^{-1} x}$
• Then $\small{f'(x)~=~\frac{1}{1 + x^2}}$

2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \tan^{-1} x~+~\rm{C}}$

Part (ii):

1. The $\small{\left[\frac{(x^2 + 1)}{(x+1)^2} \right]}$ portion can be rearranged as shown below:

$\small{\frac{x^2 + 1}{(x+1)^2} = \frac{x^2 + 1-1+1}{(x+1)^2} = \frac{x^2 -1+2}{(x+1)^2} = \frac{x^2 -1}{(x+1)^2} + \frac{2}{(x+1)^2}}$

$\small{= \frac{(x+1)(x-1)}{(x+1)^2} + \frac{2}{(x+1)^2}= \frac{x-1}{x+1} + \frac{2}{(x+1)^2}}$

2. Now, $\small{\frac{2}{(x+1)^2}}$ is the derivative of $\small{\frac{x-1}{x+1}}$.

(The reader may verify this by doing the differentiation)

3. So the given integral is of the form:

$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

Where $\small{f(x) = \frac{x-1}{x+1}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \left[\frac{x-1}{x+1} \right]~+~\rm{C}}$

---

The link below gives a few more solved examples:

Exercise 23.6

---

We have completed a discussion on integration by parts. In the next section, we will see some standard integrals.

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23.17 - Solved Examples on Integration by Parts

In the previous section, we saw the basic details about Integration by parts. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved Example 23.45
Find $\small{\int{\left[x \, \log 2x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log 2x

   ♦ Let second function be: g(x) = x

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x\right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log 2x \, \left(\frac{x^2}{2} \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{x^2}{4}}$

• This is the second term.

6. So we get:

$\small{\int{\left[x \, \log 2x \right]dx}~=~\text{First term - Second term}~=~\log 2x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}~+~\rm{C}}$

Solved Example 23.46
Find $\small{\int{\left[x^2 \, \log x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log x

   ♦ Let second function be: g(x) = $\small{x^2}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x^2\right]dx}~=~\frac{x^3}{3}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^3}{3} \right) \big]}$

• This is the first term.

4. $\small{~f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^3}{3} \right)  \big]dx}~=~\frac{x^3}{9}}$

• This is the second term.

6. So we get:

$\small{\int{\left[x^2 \, \log x \right]dx}~=~\text{First term - Second term}~=~\log x \, \left(\frac{x^3}{3} \right)~-~\frac{x^3}{9}~+~\rm{C}}$

Solved Example 23.47
Find $\small{\int{\left[x \, \sin^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sin^{-1} x}$

   ♦ Let second function be: g(x) = $\small{x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sin^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}}  \big]dx}}$

• This is the second term. But is also a product. We will do it as a sub group:

*** Beginning of sub group ***

We want $\small{\frac{1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}}  \big]dx}}$

(i) Let x = sin u

$\small{\Rightarrow \frac{dx}{du}\,=\,\cos u \Rightarrow\frac{dx}{\cos u}\,=\,du}$

Also, $\small{~\sqrt{1 - x^2}\,=\,\cos u}$

(ii) So we want:

$\small{\frac{1}{2}\int{\big[\frac{\sin^2 u}{\cos u}  \big]dx}\,=\,\frac{1}{2}\int{\big[\sin^2 u  \big]du}}$

(iii) This can be calculated as:

$\small{\frac{1}{2}\int{\big[\sin^2 u  \big]du}\,=\,\frac{1}{2}\int{\big[\frac{1 - \cos 2u}{2}  \big]du}\,=\,\frac{1}{4}\int{\big[1 - \cos 2u  \big]du}}$

$\small{~=~\frac{u}{4}\,-\,\frac{\sin 2u}{8}~=~\frac{u}{4}\,-\,\frac{2 \sin u \cos u}{8}}$

$\small{~=~\frac{u}{4}\,-\,\frac{\sin u \cos u}{4}~=~\frac{\sin^{-1}x}{4}\,-\,\frac{x \sqrt{1 - x^2}}{4}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x \, \sin^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\sin^{-1}x \, \left(\frac{x^2}{2}\right) \big]~-~\big[\frac{\sin^{-1}x}{4}\,-\,\frac{x \sqrt{1 - x^2}}{4} \big]}$

$\small{~=~\sin^{-1}x \, \left(\frac{x^2}{2}\,-\,\frac{1}{4}\right)\,+\,\frac{x \sqrt{1 - x^2}}{4}\,+\,\rm{C}}$

Solved Example 23.48
Find $\small{\int{\left[x \, \tan^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\tan^{-1} x}$

   ♦ Let second function be: g(x) = $\small{x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\tan^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{1+x^2}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{1+x^2}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{1}{2}\int{\big[\frac{x^2}{1+x^2}  \big]dx}}$

• This is the second term. But is also a product. We will do it as a sub group:

*** Beginning of sub group ***

We want $\small{\frac{1}{2}\int{\big[\frac{x^2}{1+x^2}  \big]dx}}$

(i) Let x = tan u

$\small{\Rightarrow \frac{dx}{du}\,=\,\sec^2 u \Rightarrow\frac{dx}{\sec^2 u}\,=\,du}$

Also, $\small{~1 + x^2\,=\,\sec^2 u}$

(ii) So we want:

$\small{\frac{1}{2}\int{\big[\frac{\tan^2 u}{\sec^2 u}  \big]dx}\,=\,\frac{1}{2}\int{\big[\tan^2 u  \big]du}}$

(iii) This can be calculated as:

$\small{\frac{1}{2}\int{\big[\tan^2 u  \big]du}\,=\,\frac{1}{2}\int{\big[\sec^2 u \,-\,1  \big]du}\,=\,\frac{\tan u}{2}\,-\,\frac{u}{2}}$

$\small{\,=\,\frac{x}{2}\,-\,\frac{\tan^{-1}x}{2}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x \, \tan^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\tan^{-1}x \, \left(\frac{x^2}{2}\right) \big]~-~\big[\frac{x}{2}\,-\,\frac{\tan^{-1}x}{2}\big]}$

$\small{~=~\tan^{-1}x \, \left(\frac{x^2}{2}\,+\,\frac{1}{2}\right)\,-\,\frac{x}{2}\,+\,\rm{C}}$

Solved Example 23.49
Find $\small{\int{\left[x \, \cos^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\cos^{-1} x}$

   ♦ Let second function be: g(x) = $\small{x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\cos^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-1}{\sqrt{1-x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{-1}{\sqrt{1-x^2}}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{-1}{2}\int{\big[\frac{1}{\sqrt{1-x^2}}  \big]dx}}$

• This is the second term. But is also a product. We will do it as a sub group:

*** Beginning of sub group ***

We want $\small{\frac{-1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}}  \big]dx}}$

(i) Let x = cos u

$\small{\Rightarrow \frac{dx}{du}\,=\,-\sin u \Rightarrow\frac{-dx}{\sin u}\,=\,du}$

Also, $\small{~\sqrt{1 - x^2}\,=\,\sin u}$

(ii) So we want:

$\small{\frac{-1}{2}\int{\big[\frac{\cos^2 u}{\sin u}  \big]dx}\,=\,\frac{1}{2}\int{\big[\cos^2 u  \big]du}}$

(iii) This can be calculated as:

$\small{\frac{1}{2}\int{\big[\cos^2 u  \big]du}\,=\,\frac{1}{2}\int{\big[\frac{1 + \cos 2u}{2}  \big]du}\,=\,\frac{1}{4}\int{\big[1 + \cos 2u  \big]du}}$

$\small{~=~\frac{u}{4}\,+\,\frac{\sin 2u}{8}~=~\frac{u}{4}\,+\,\frac{2 \sin u \cos u}{8}}$

$\small{~=~\frac{u}{4}\,+\,\frac{\sin u \cos u}{4}~=~\frac{\cos^{-1}x}{4}\,+\,\frac{x \sqrt{1 - x^2}}{4}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x \, \cos^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\cos^{-1}x \, \left(\frac{x^2}{2}\right) \big]~-~\big[\frac{\cos^{-1}x}{4}\,+\,\frac{x \sqrt{1 - x^2}}{4} \big]}$

$\small{~=~\cos^{-1}x \, \left(\frac{x^2}{2}\,-\,\frac{1}{4}\right)\,-\,\frac{x \sqrt{1 - x^2}}{4}\,+\,\rm{C}}$

Solved Example 23.50
Find $\small{\int{\left[\sin^{-1}x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sin^{-1} x}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sin^{-1}x \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(x \right)  \big]dx}~=~-\sqrt{1-x^2}}$

• This is second term.

(The reader may write all steps involved in this integration process)

6. So we get:

$\small{\int{\left[\sin^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\sin^{-1}x \, \left(x\right) \big]~-~\big[-\sqrt{1-x^2} \big]}$

$\small{~=~x \sin^{-1}x~+~\sqrt{1-x^2}\,+\,\rm{C}}$

Solved Example 23.51
Find $\small{\int{\left[(\sin^{-1}x)^2 \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sin^{-1} x}$

   ♦ Let second function be: g(x) = $\small{\sin^{-1}x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin^{-1}x \right]dx}~=~x \sin^{-1}x~+~\sqrt{1-x^2}}$

(See solved example 23.50 above)

3. $\small{\big[f(x) \left(A \right) \big]~=~\sin^{-1}x \big[x \sin^{-1}x~+~\sqrt{1-x^2} \big]}$

$\small{~=~\big[x (\sin^{-1}x)^2~+~\sin^{-1}x\,  \left(\sqrt{1-x^2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(x \sin^{-1}x~+~\sqrt{1-x^2} \right)  \big]dx}}$

$\small{~=~\int{\big[\frac{x \sin^{-1}x}{\sqrt{1-x^2}}~+~1  \big]dx}~=~x\,-\,(\sqrt{1-x^2}) \sin^{-1}x~+~x}$

$\small{~=~2x\,-\,\left(\sqrt{1-x^2} \right) \sin^{-1}x}$

(See solved example 23.38 of the previous section)

• This is second term.

6. So we get:

$\small{\int{\left[(\sin^{-1} x)^2 \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x (\sin^{-1}x)^2~+~\sin^{-1}x\,  \left(\sqrt{1-x^2}\right) \big]~-~\big[2x\,-\,\left(\sqrt{1-x^2} \right) \sin^{-1}x \big]}$

$\small{~=~x (\sin^{-1}x)^2~+~2 \sin^{-1}x\,  \left(\sqrt{1-x^2}\right) ~-~2x}$

Solved Example 23.52
Find $\small{\int{\left[\frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\cos^{-1}x}$

   ♦ Let second function be: g(x) = $\small{\frac{x }{\sqrt{1 - x^2}} }$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\frac{x }{\sqrt{1 - x^2}} \right]dx}~=~-\sqrt{1-x^2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[(-1)(\cos^{-1}x) \,\sqrt{1-x^2} \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-1}{\sqrt{1 - x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{-1}{\sqrt{1 - x^2}}\,\left(-\sqrt{1-x^2} \right)  \big]dx}~=~x}$

• This is the second term.

6. So we get:

$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[(-1)(\cos^{-1}x) \,\sqrt{1-x^2} \big]~-~\big[x \big]~+~\rm{C}}$

$\small{~=~- \big[x~+~(\cos^{-1}x) \,\sqrt{1-x^2}\big]~+~\rm{C}}$

Solved Example 23.53
Find $\small{\int{\left[x \sec^2 x \right]dx}}$
Solution:
$\small{\int{\left[x \sec^2 x \right]dx}}$

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{x}$

   ♦ Let second function be: g(x) = $\small{\sec^2 x }$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sec^2 x \right]dx}~=~\tan x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \tan x \big]}$

• This is the first term.

4. $\small{f'(x)~=~1}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[1\,\left(\tan x \right)  \big]dx}~=~\log \left|\sec x \right| }$

• This is the second term.

6. So we get:

$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \tan x \big]~-~\big[\log \left|\sec x \right| \big]~+~\rm{C}}$

$\small{~=~x \tan x ~+~\log \left|\frac{1}{\sec x} \right| ~+~\rm{C}}$

$\small{~=~x \tan x ~+~\log \left|\cos x \right| ~+~\rm{C}}$

Solved Example 23.54
Find $\small{\int{\left[\tan^{-1}x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\tan^{-1} x}$

   ♦ Let second function be: g(x) = $\small{1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\tan^{-1}x \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{1+x^2}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{1+x^2}\,\left(x \right)  \big]dx}~=~\frac{1}{2}\log \left|1+x^2 \right|}$

• This is second term.

(The reader may write all steps involved in this integration process)

6. So we get:

$\small{\int{\left[\tan^{-1} x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\tan^{-1}x \, \left(x\right) \big]~-~\big[\frac{1}{2}\log \left|1+x^2 \right| \big]}$

$\small{~=~x \tan^{-1}x~-~\frac{1}{2}\log \left|1+x^2 \right|\,+\,\rm{C}}$

Solved Example 23.55
Find $\small{\int{\left[x (\log x)^2 \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{(\log x)^2}$

   ♦ Let second function be: g(x) = $\small{x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[(\log x)^2 \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{2 \log \left|x \right|}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{2 \log \left|x \right|}{x}\,\left(\frac{x^2}{2} \right)  \big]dx}}$

$\small{~=~\int{\big[x\log \left|x \right|  \big]dx}~=~\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}}$

(See solved example 23.44 of the previous section)

• This is second term.

6. So we get:

$\small{\int{\left[x (\log x)^2 \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[(\log x)^2 \, \left(\frac{x^2}{2}\right) \big]~-~\big[\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4} \big]}$

$\small{~=~(\log x)^2 \, \left(\frac{x^2}{2}\right) ~-~\log x \, \left(\frac{x^2}{2} \right)~+~\frac{x^2}{4} }$

Solved Example 23.56
Find $\small{\int{\left[(x^2 + 1) \log x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log x

   ♦ Let second function be: g(x) = $\small{x^2 + 1}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x^2 + 1\right]dx}~=~\frac{x^3}{3}~+~x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^3}{3}~+~x \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^3}{3}~+~x \right)  \big]dx}~=~\frac{x^3}{9}~+~x}$

• This is the second term.

6. So we get:

$\small{\int{\left[(x^2 + 1) \log x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\log x \, \left(\frac{x^3}{3}~+~x \right)~-~\frac{x^3}{9}~-~x~+~\rm{C}}$

---

We have seen the method of integration by parts. In the next section, we will see a special case in this method.

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23.16 - Integration by Parts

In the previous section, we completed a discussion on integration by partial fractions. In this section, we will see the third method for integration, which is: Integration by parts.

This method is useful for integrating "products of functions". It can be explained in 6 steps:

1. Consider the product rule:

$\small{\frac{d}{dx}(uv)~=~u \frac{dv}{dx}\,+\,v \frac{du}{dx}}$

2. Integrating both sides, we get:

$\small{uv~=~\int{\left[u \frac{dv}{dx} \right]dx}\,+\,\int{\left[v \frac{du}{dx} \right]dx}}$

⇒ $\small{\int{\left[u \frac{dv}{dx} \right]dx}~=~uv\,-\,\int{\left[v \frac{du}{dx} \right]dx}}$

Note: In the above step, there will be a constant of integration. But as we continue to write more steps, there will be more constants. We just need to combine all those constants into a single constant. This can be done in the final step.

3. Let $\small{u = f(x) ~~\rm{and}~~\frac{dv}{dx} = g(x)}$

4. Then we can write:

$\small{\frac{du}{dx} = f'(x) ~~\rm{and}~~v = \int{\left[g(x) \right]dx}}$

5. Now the result in (2) becomes:

$\small{\int{\left[f(x)\, g(x) \right]dx}~=~\big[f(x) \left(\int{\left[g(x) \right]dx} \right) \big]\,-\,\int{\big[\left(\int{\left[g(x) \right]dx} \right)\,f'(x)  \big]dx}}$

⇒ $\small{\int{\left[f(x)\, g(x) \right]dx}~=~\big[f(x) \left(\int{\left[g(x) \right]dx} \right) \big]\,-\,\int{\big[f'(x)\,\left(\int{\left[g(x) \right]dx} \right)\,  \big]dx}}$

⇒ $\small{\int{\left[f(x)\, g(x) \right]dx}~=~\big[f(x) \left(A \right) \big]\,-\,\int{\big[f'(x)\,\left(A \right)\,  \big]dx}~+~\rm{C}}$

Where $\small{A~=~\int{\left[g(x) \right]dx} }$

6. Let us compile the above steps:

• We are asked to integrate a function, which is the product of two functions.

• Let f(x) be the first function and g(x) be the second function.

The compiled form can be written in three steps:

(i) Find $\small{\big[f(x) \left(A \right) \big]}$. This is the first term.

(ii) Find $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}}$. This is the second term.

(iii) Subtract second term from first term and add the constant of integration.

---

Let us see a solved example:

Solved Example 23.35
Find $\small{\int{\left[x \cos x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = x

   ♦ Let second function be: g(x) = cos x

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\cos x \right]dx}~=~\sin x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \, \sin x \big]}$

• This is the first term.

4. $\small{f'(x)~=~1}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[1\,\left(\sin x \right)  \big]dx}~=~-\cos x}$

• This is the second term.

6. So we get:

$\small{\int{\left[x \cos x \right]dx}~=~\text{First term - Second term}~=~x \sin x ~+~\cos x~+~\rm{C}}$

---

In the above solved example, let us interchange f(x) and g(x). We will write the steps again:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = cos x

   ♦ Let second function be: g(x) = x

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x \right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\cos x \, \left(\frac{x^2}{2} \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~- \sin x}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[- \sin x\,\left(\frac{x^2}{2} \right)  \big]dx}}$

• This is the second term.

6. We see that:

The second term is the integral of yet another product, also with higher power of x. That means, the calculations are becoming lengthy.

• So it is clear that, selection of f(x) and g(x) is important.

---

The following three rules can be used to select the first function f(x):
(i) One factor is a polynomial function.
• The other factor is a function other than inverse trigonometric or logarithmic.
• Then the polynomial function should be taken as f(x).

(ii) One factor is a polynomial function.
• The other factor is an inverse trigonometric function.
• Then the inverse trigonometric function should be taken as f(x).

(iii) One factor is a polynomial function.
• The other factor is a logarithmic function.
• Then the logarithmic function should be taken as f(x).

(iv) In general, we must take that function as the second function, whose integral is well known to us.

---

In the above solved example, we do integration two times. First in step (2) and later in step (5). But we add the constant of integration only in the final step. Let us see if it makes any difference when the constant is added in every integration step. We will write all the steps again:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = x

   ♦ Let second function be: g(x) = cos x

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\cos x \right]dx}~=~\sin x~+~\rm{C_1}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \,\left( \sin x~+~\rm{C_1}\right) \big]~=~\big[x \, \sin x~+~x\,\rm{C_1} \big]}$

• This is the first term.

4. $\small{f'(x)~=~1}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[1\,\left(\sin x~+~\rm{C_1} \right)  \big]dx}~=~-\cos x~+~\rm{C_1}x~+~\rm{C_2}}$

• This is the second term.

6. So we get:

$\small{\int{\left[x \cos x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \, \sin x~+~x\,\rm{C_1} \big]~-~\left[-\cos x~+~\rm{C_1}x~+~\rm{C_2} \right]}$

$\small{~=~x \, \sin x~+~x\,\rm{C_1} ~+~\cos x~-~\rm{C_1}x~-~\rm{C_2} }$

$\small{~=~x \, \sin x ~+~\cos x~-~\rm{C_2} }$

$\small{~=~x \, \sin x ~+~\cos x~+~\rm{C} }$

◼ It is clear that, adding constants in the intermediate steps will not make any difference. We need to add a constant in the final step only.

---

Solved Example 23.36
Find $\small{\int{\left[x e^x \right]dx}}$
Solution:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = x

   ♦ Let second function be: g(x) = $\small{e^x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^x \right]dx}~=~e^x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \,e^x \big]}$

• This is the first term.

4. $\small{f'(x)~=~1}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[1\,\left(e^x \right)  \big]dx}~=~e^x}$

• This is the second term.

6. So we get:

$\small{\int{\left[x\, e^x \right]dx}~=~\text{First term - Second term}~=~x \, e^x~-~e^x~+~\rm{C}}$

Solved Example 23.37
Find $\small{\int{\left[x e^{6x} \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = x

   ♦ Let second function be: g(x) = $\small{e^{6x}}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^{6x} \right]dx}~=~\frac{e^{6x}}{6}}$

(The reader may write all steps involved in the integration process)

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \, \left(\frac{e^{6x}}{6} \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~1}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[1\,\left(\frac{e^{6x}}{6} \right)  \big]dx}~=~\frac{e^{6x}}{36}}$

(The reader may write all steps involved in the integration process)

• This is the second term.

6. So we get:

$\small{\int{\left[x\, e^{6x} \right]dx}~=~\text{First term - Second term}~=~x \, \left(\frac{e^{6x}}{6} \right)~-~\frac{e^{6x}}{36}~+~\rm{C}}$

Solved Example 23.38
Find $\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sin^{-1}x}$

   ♦ Let second function be: g(x) = $\small{\frac{x }{\sqrt{1 - x^2}} }$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\frac{x }{\sqrt{1 - x^2}} \right]dx}~=~-\sqrt{1-x^2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[(-1)(\sin^{-1}x) \,\sqrt{1-x^2} \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1 - x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{\sqrt{1 - x^2}}\,\left(-\sqrt{1-x^2} \right)  \big]dx}~=~-x}$

• This is the second term.

6. So we get:

$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[(-1)(\sin^{-1}x) \,\sqrt{1-x^2} \big]~-~\big[-x \big]~+~\rm{C}}$

$\small{~=~x~-~(\sin^{-1}x) \,\sqrt{1-x^2}~+~\rm{C}}$

Alternate method:

• Put $\small{\sin^{-1}x = \theta}$

• Then we get: $\small{\sin \theta = x}$

• Also, $\small{\frac{d \theta}{dx}~=~\frac{1}{\sqrt{1 -x^2}}}$

$\small{\Rightarrow~\frac{dx}{\sqrt{1 - x^2}}~=~d\theta}$

• So the question becomes: $\small{\int{\left[\sin \theta\,(\theta)\right]dx}}$

Now we can write the usual steps:

1. Assigning first and second functions:

   ♦ Let first function be: $\small{f(\theta)~=~\theta}$

   ♦ Let second function be: $\small{g(\theta)~=~\sin \theta}$

2. Finding A:

$\small{A~=~\int{\left[g(\theta) \right]dx}~=~\int{\left[\sin \theta \right]dx}~=~-\cos \theta}$

3. $\small{\big[f(\theta) \left(A \right) \big]~=~\big[(-1)(\theta) \,\cos \theta \big]}$

• This is the first term.

4. $\small{f'(\theta)~=~1}$

5. $\small{\int{\big[f'(\theta)\,\left(A \right)  \big]d\theta}~=~\int{\big[1\,\left(-\cos \theta \right)  \big]d\theta}~=~-\sin \theta}$

• This is the second term.

6. So we get:

$\small{\int{\left[\sin \theta\,(\theta) \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\sin \theta~-~\theta \cos \theta~+~\rm{C}}$

7. Substituting for $\small{\theta}$, we get:

$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~x~-~(\sin^{-1} x)\sqrt{1 - x^2}~+~\rm{C}}$

 

Solved Example 23.39
Find $\small{\int{\left[\log x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log x

   ♦ Let second function be: g(x) = 1

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1 \right]dx}~=~ x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, (1) \big]~=~\big[\log x \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(x \right)  \big]dx}~=~ x}$

• This is the second term.

6. So we get:

$\small{\int{\left[\log x \right]dx}~=~\text{First term - Second term}~=~\log x ~-~ x~+~\rm{C}}$

Solved Example 23.40
Find $\small{\int{\left[e^x \, \sin x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{e^x}$

   ♦ Let second function be: g(x) = $\small{\sin x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin x \right]dx}~=~-\cos x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[e^x \, \left(-\cos x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~e^x}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[e^x\,\left(-\cos x \right)  \big]dx}~=~(-1)\int{\big[e^x\,\left(\cos x \right)  \big]dx}}$

• This is similar to the given question. We will do it as a sub group:

*** Beginning of sub group ***

(i) Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{e^x}$

   ♦ Let second function be: g(x) = $\small{\cos x}$

(ii) Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\cos x \right]dx}~=~\sin x}$

(iii) $\small{\big[f(x) \left(A \right) \big]~=~\big[e^x \, \left(\sin x\right) \big]}$

• This is the first term.

(iv) $\small{f'(x)~=~e^x}$

(v) $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[e^x\,\left(\sin x \right)  \big]dx}}$

• This is the second term.

(vi) So we get:

$\small{\int{\big[e^x\,\left(\cos x \right)  \big]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[e^x \, \left(\sin x\right) \big]~-~\int{\big[e^x\,\left(\sin x \right)  \big]dx}~+~\rm{C}}$

$\small{\Rightarrow (-1)\int{\big[e^x\,\left(\cos x \right)  \big]dx}~=~\int{\big[e^x\,\left(\sin x \right)  \big]dx}~-~\big[e^x \, \left(\sin x\right) \big]~-~\rm{C}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[e^x \, \sin x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[e^x \, \left(-\cos x\right) \big]~-~\Bigg[\int{\big[e^x\,\left(\sin x \right)  \big]dx}~-~\big[e^x \, \left(\sin x\right) \big]~-~\rm{C} \Bigg]}$

$\small{~=~-\big[e^x \, \left(\cos x\right) \big]~-~\int{\big[e^x\,\left(\sin x \right)  \big]dx}~+~\big[e^x \, \left(\sin x\right) \big]~+~\rm{C} }$

$\small{\Rightarrow 2\int{\big[e^x\,\left(\sin x \right)  \big]dx}~=~-\big[e^x \, \left(\cos x\right) \big]~+~\big[e^x \, \left(\sin x\right) \big]~+~\rm{C} }$

$\small{\Rightarrow \int{\big[e^x\,\left(\sin x \right)  \big]dx}~=~\frac{e^x (\sin x \,-\, \cos x)}{2}~+~\rm{C} }$

Solved Example 23.41
Find $\small{\int{\left[x \sin x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = x

   ♦ Let second function be: g(x) = $\small{\sin x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin x \right]dx}~=~-\cos x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \, \left(-\cos x \right) \big]}$

• This is the first term.

4. $\small{~f'(x)~=~1}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[1\,\left(-\cos x \right)  \big]dx}~=~-\sin x}$

• This is the second term.

6. So we get:

$\small{\int{\left[x\, \sin x \right]dx}~=~\text{First term - Second term}~=~\big[x \, \left(-\cos x \right) \big]~-~[-\sin x]~+~\rm{C}}$

$\small{~=~-x \cos x ~+~\sin x~+~\rm{C}}$

Solved Example 23.42
Find $\small{\int{\left[x \sin 3x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = x

   ♦ Let second function be: g(x) = $\small{\sin 3x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin 3x \right]dx}~=~-\frac{\cos 3x}{3}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \, \left(-\frac{\cos 3x}{3} \right) \big]~=~ -\frac{x\cos 3x}{3}}$

• This is the first term.

4. $\small{~f'(x)~=~1}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[1\,\left(-\frac{\cos 3x}{3} \right)  \big]dx}~=~-\frac{\sin 3x}{9}}$

• This is the second term.

6. So we get:

$\small{\int{\left[x\, \sin 3x \right]dx}~=~\text{First term - Second term}~=~\big[-\frac{x \cos 3x}{3} \big]~-~\left[-\frac{\sin 3x}{9}\right]~+~\rm{C}}$

$\small{~=~-\frac{x\cos 3x}{3} ~+~\frac{\sin 3x}{9}~+~\rm{C}}$

Solved Example 23.43
Find $\small{\int{\left[x^2 e^x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{x^2}$

   ♦ Let second function be: g(x) = $\small{e^x}$

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^x \right]dx}~=~e^x}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x^2 \, \left(e^x \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~2x}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[2x\,\left(e^x \right)  \big]dx}}$

• This is the second term. We will do it as a sub group:

*** Beginning of sub group ***

(i) Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{2x}$

   ♦ Let second function be: g(x) = $\small{e^x}$

(ii) Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[e^x\right]dx}~=~e^x}$

(iii) $\small{\big[f(x) \left(A \right) \big]~=~\big[2x \, \left(e^x\right) \big]}$

• This is the first term.

(iv) $\small{f'(x)~=~2}$

(v) $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[2\,\left(e^x \right)  \big]dx}~=~2 e^x}$

• This is the second term.

(vi) So we get:

$\small{\int{\big[2x\,e^x  \big]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[2x \,e^x \big]~-~\big[2e^x  \big]}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x^2\,e^x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x^2 \, \left(e^x \right) \big]~-~\Bigg[\big[2x \,e^x \big]~-~\big[2e^x  \big] \Bigg]~+~\rm{C}}$

$\small{~=~e^x(x^2~-~2x~+~2)~+~\rm{C}}$

Solved Example 23.44
Find $\small{\int{\left[x \, \log x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log x

   ♦ Let second function be: g(x) = x

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x\right]dx}~=~\frac{x^2}{2}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^2}{2} \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{x^2}{4}}$

• This is the second term.

6. So we get:

$\small{\int{\left[x \, \log x \right]dx}~=~\text{First term - Second term}~=~\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}~+~\rm{C}}$

---

In the next section, we will see a few more solved examples.

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23.15 - More Solved Examples on Integration by Partial Fractions

In the previous section, we saw some solved examples on integration by partial fractions. In this section, we will see some solved examples which involve both substitution and partial fractions.

Solved Example 23.30
Find $\small{\int{\left[\frac{\cos x}{\sin^2 x - \sin x} \right]dx} }$
Solution:
1. Put u = sin x. Then du/dx = cos x
⇒ cos x dx = du

• So we want:
$\small{\int{\left[\frac{\cos x}{\sin^2 x - \sin x} \right]dx}~=~\int{\left[\frac{1}{u^2 - u} \right]du}}$

• The numerator is a polynomial of degree zero. The denominator is also a polynomial of degree 2.
2. So it is a proper rational function. We can straight away start partial fraction decomposition

3. First we factorize the denominator. We get:
u² − u = u(u−1)
   ♦ All factors are linear
   ♦ And all factors are distinct from one another.

4. So this is case I. We are able to write:

$\small{\frac{1}{u^2 - u}\,=\,\frac{1}{u(u-1)}\,=\,\frac{A_1}{u}~+~\frac{A_2}{u-1}}$

Where A₁ and A₂ are real numbers.

5. To find A₁ and A₂, we make denominators same on both sides:

$\small{\frac{1}{u(u-1)}\,=\,\frac{A_1}{u}~+~\frac{A_2}{u-1}~=~\frac{A_1 (u-1)~+~A_2 (u)}{u(u-1)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:

$\small{1~=~A_1 (u-1)~+~A_2 (u)}$

7. After equating the numerators, we can use suitable substitution.

♦ Put u = 1. We get: A₂ = 1

♦ Put u = 0. We get: A₁ = −1

8. Now the result in (4) becomes:

$\small{\frac{1}{u^2 - u}\,=\,\frac{1}{u(u-1)}\,=\,\frac{-1}{u}~+~\frac{1}{u-1}}$

9. So the integration becomes easy. We get:

$\small{- \log \left|u \right| + \log \left|u-1 \right|~+~\rm{C}}$

• The reader may write all the steps involved in the integration process.

10. Substituting for u, we get:

$\small{- \log \left|\sin x \right| + \log \left|\sin x-1 \right|~+~\rm{C}}$

Solved Example 23.31
Find $\small{\int{\left[\frac{(3 \sin x - 2)\cos x}{5 - \cos^2 x - 4 \sin x} \right]dx} }$
Solution:
1. Put u = sin x. Then du/dx = cos x
⇒ cos x dx = du

• So we want:

$\small{\int{\left[\frac{(3 \sin x - 2)\cos x}{5 - \cos^2 x - 4 \sin x} \right]dx}~=~\int{\left[\frac{(3 \sin x - 2)\cos x}{5 - (1 - \sin^2 x) - 4 \sin x} \right]dx}~=~\int{\left[\frac{3u - 2}{4 + u^2 - 4u} \right]du}}$

• The numerator is a polynomial of degree 1. The denominator is  a polynomial of degree 2.

2. So it is a proper rational function. We can straight away start partial fraction decomposition

3. First we factorize the denominator. We get:

4 + u² − 4u = (u−2)²
   ♦ All factors are linear
   ♦ And all factors are not distinct from one another.

4. So this is case II. We are able to write:

$\small{\frac{3u - 2}{4 + u^2 - 4u}\,=\,\frac{3u - 2}{(u-2)^2}\,=\,\frac{A_1}{u-2}~+~\frac{A_2}{(u-2)^2}}$

Where A₁ and A₂ are real numbers.
5. To find A₁ and A₂, we make denominators same on both sides:

$\small{\frac{3u - 2}{(u-2)^2}\,=\,\frac{A_1}{u-2}~+~\frac{A_2}{(u-2)^2}\,=\,\frac{A_1(u-2)~+~A_2}{(u-2)^2}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:

$\small{3u - 2~=~A_1 (u-2)~+~A_2}$

7. After equating the numerators, we can use suitable substitution.

♦ Put u = 2. We get: 4 = A₂

♦ Put u = 0. We get: −2 = −2 A₁ + 4. So A₁ = 3

8. Now the result in (4) becomes:

$\small{\frac{3u - 2}{4 + u^2 - 4u}\,=\,\frac{3}{u-2}~+~\frac{4}{(u-2)^2}}$

9. So the integration becomes easy. We get:

$\small{3\log \left|u-2 \right| - \frac{4}{u-2}~+~\rm{C}}$

• The reader may write all the steps involved in the integration process.

10. Substituting for u, we get:

$\small{3\log \left|\sin x-2 \right| - \frac{4}{\sin x-2}~+~\rm{C}}$

$\small{~=~3\log \left|\sin x-2 \right| + \frac{4}{2 - \sin x}~+~\rm{C}}$

$\small{~=~3\log (2 - \sin x)  + \frac{4}{2 - \sin x}~+~\rm{C}}$

Since (2 - sin x) is always +ve.

Solved Example 23.32

Find $\small{\int{\left[\frac{\cos x}{(1 - \sin x)(2 - \sin x)} \right]dx}}$
[Hint: Put sin x = u]
Solution:
1. Put u = sin x. Then du/dx = cos x
⇒ cos x dx = du

• So we want:

$\small{\int{\left[\frac{\cos x}{(1 - \sin x)(2 - \sin x)}\right]dx}~=~\int{\left[\frac{1}{(1-u)(2-u)} \right]du}}$

• The numerator is a polynomial of degree zero. The denominator is  a polynomial of degree 2.

2. So it is a proper rational function. We can straight away start partial fraction decomposition

3. First we factorize the denominator. But it is already in the factorized form:

(1−u)(2−u)
   ♦ All factors are linear
   
   ♦ And all factors are distinct from one another.

4. So this is case I. We are able to write:

$\small{\frac{1}{(1-u)(2-u)}\,=\,\frac{A_1}{1-u}~+~\frac{A_2}{2-u}}$

Where A₁ and A₂ are real numbers.

5. To find A₁ and A₂, we make denominators same on both sides:

$\small{\frac{1}{(1-u)(2-u)}\,=\,\frac{A_1}{1-u}~+~\frac{A_2}{2-u}\,=\,\frac{A_1(2-u)~+~A_2(1-u)}{(1-u)(2-u)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:

$\small{1~=~A_1(2-u)~+~A_2(1-u)}$

7. After equating the numerators, we can use suitable substitution.

♦ Put u = 2. We get: A₂ = −1

♦ Put u = 1. We get: A₁ = 1

8. Now the result in (4) becomes:

$\small{\frac{1}{(1-u)(2-u)}\,=\,\frac{1}{1-u}~-~\frac{1}{2-u}}$

9. So the integration becomes easy. We get:

$\small{-\log \left|1-u \right| + \log \left|2-u \right|~+~\rm{C}}$

• The reader may write all the steps involved in the integration process.

10. Substituting for u, we get:

$\small{-\log \left|1-\sin x \right| + \log \left|2-\sin x \right|~+~\rm{C}}$

$\small{~=~\log \left| \frac{2 - \sin x}{1 - \sin x} \right|~+~\rm{C}}$

Alternate method:

1. Put t = sin x. Then dt/dx = cos x
⇒ cos x dx = dt

• So we want:

$\small{\int{\left[\frac{\cos x}{(1 - \sin x)(2 - \sin x)}\right]dx}~=~\int{\left[\frac{dt}{(1-t)(2-t)} \right]}~=~\int{\left[\frac{dt}{2 - 3t + t^2} \right]}}$

$\small{~=~\int{\left[\frac{dt}{t^2 - 3t + 2} \right]}}$

2. Recall how we analyzed formula VII: $\small{\int{\left[\frac{dt}{a t^2\,+\,b t \,+\,c} \right]}~=~\frac{1}{a}\int{\left[\frac{dx}{u^2~\pm~ k^2} \right]}}$

• In our present case, a = 1, b = −3 and c = 2

3. So we can calculate u and k²:

• $\small{u\,=\,t\,+\,\frac{b}{2a}~=~t\,+\,\frac{-3}{2(1)}~=~(t-3/2)}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{2}{1}\,-\,\frac{(-3)^2}{4(1)^2}~=~2\,-\,\frac{9}{4}~=~-(1/4)}$

4. So we want:

$\small{\int{\left[\frac{dt}{t^2\,-\,3x\,+\,2} \right]}~=~\frac{1}{a}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

$\small{~=~\frac{1}{1}\int{\left[\frac{dt}{(t-3/2)^2~-~(1/2)^2} \right]}}$

[Recall that, we put u = t + b/(2a). So du = dt]

5. This integration can be done as shown below:

(i) Put v = (t−3/2). Then dv/dt = 1, which gives dv = dt

• So we want:

$\small{\int{\left[\frac{dt}{(t-3/2)^2~-~(1/2)^2} \right]}~=~\int{\left[\frac{dv}{v^2\,-\,(1/2)^2} \right]}}$

(ii) We have formula I: $\small{\int{\left[\frac{dv}{v^2\,-\,m^2} \right]}\,=\, \frac{1}{2m} \log \left| \frac{v-m}{v+m} \right|\,+\,\rm{C}}$

• In our present case, m = 1/2

6. So we get:
$\small{\int{\left[\frac{dv}{v^2\,-\,(1/2)^2}\right]}\,=\, \frac{1}{2(1/2)} \log \left| \frac{v - (1/2)}{v + (1/2)} \right|\,+\,C}$

$\small{\,=\,  \log \left| \frac{t - (3/2) - (1/2)}{t - (3/2) + (1/2)} \right|\,+\,C\,=\,  \log \left| \frac{t - 2}{t - 1} \right|\,+\,C\,=\,  \log \left| \frac{\sin x - 2}{\sin x - 1} \right|\,+\,C}$

Solved Example 23.33
Find $\small{\int{\left[\frac{1}{e^x - 1} \right]dx}}$
[Hint: Put e^x = u]
Solution:
1. Put u = e^x. Then du/dx = e^x
⇒ e^x dx = du

• So we want:

$\small{\int{\left[\frac{1}{e^x - 1}\right]dx}~=~\int{\left[\frac{e^x}{e^x(e^x - 1)}\right]dx}~=~\int{\left[\frac{1}{u(u-1)} \right]du}}$

• The numerator is a polynomial of degree zero. The denominator is  a polynomial of degree 2.

2. So it is a proper rational function. We can straight away start partial fraction decomposition

3. First we factorize the denominator. But it is already in the factorized form:

u(u−1)

   ♦ All factors are linear
   
   ♦ And all factors are distinct from one another.

4. So this is case I. We are able to write:

$\small{\frac{1}{u(u-1)}\,=\,\frac{A_1}{u}~+~\frac{A_2}{u-1}}$

Where A₁ and A₂ are real numbers.

5. To find A₁ and A₂, we make denominators same on both sides:

$\small{\frac{1}{u(u-1)}\,=\,\frac{A_1}{u}~+~\frac{A_2}{u-1}\,=\,\frac{A_1(u-1)~+~A_2 u}{u(u-1)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:

$\small{1~=~A_1(u-1)~+~A_2 u}$

7. After equating the numerators, we can use suitable substitution.

♦ Put u = 1. We get: A₂ = 1

♦ Put u = 0. We get: A₁ = −1

8. Now the result in (4) becomes:

$\small{\frac{1}{u(u-1)}\,=\,\frac{-1}{u}~+~\frac{1}{u-1}}$

9. So the integration becomes easy. We get:

$\small{-\log \left|u \right| + \log \left|u-1 \right|~+~\rm{C}}$

• The reader may write all the steps involved in the integration process.

10. Substituting for u, we get:

$\small{-\log \left|e^x \right| + \log \left|e^x - 1 \right|~+~\rm{C}}$

$\small{~=~\log \left| \frac{e^x - 1}{e^x} \right|~+~\rm{C}}$

Solved Example 23.34
Find $\small{\int{\left[\frac{1}{x(x^n + 1)} \right]dx}}$
[Hint: Multiply numerator and denominator by x^n-1 and put xⁿ = u]
Solution:
1. We have:

$\small{\int{\left[\frac{1}{x(x^n + 1)} \right]dx}~=~\int{\left[\frac{x^{n-1}}{x^{n-1}\,x(x^n + 1)} \right]dx}~=~\int{\left[\frac{x^{n-1}}{x^{n}(x^n + 1)} \right]dx}~=~\int{\left[\frac{n\,x^{n-1}}{n\,x^{n}(x^n + 1)} \right]dx}}$

• Put $u = x^n$. Then $\small{\frac{du}{dx}~=~n x^{n-1}}$

⇒ $\small{n x^{n-1}\,dx}~=~du$

• So we want:

$\small{\int{\left[\frac{1}{x(x^n + 1)} \right]dx}~=~\int{\left[\frac{n\,x^{n-1}}{n\,x^{n}(x^n + 1)} \right]dx}~=~\int{\left[\frac{1}{nu(u+1)} \right]du}}$

• The numerator is a polynomial of degree zero. The denominator is  a polynomial of degree 2.

2. So it is a proper rational function. We can straight away start partial fraction decomposition

3. First we factorize the denominator. But it is already in the factorized form:

u(u+1)

   ♦ All factors are linear
   
   ♦ And all factors are distinct from one another.

4. So this is case I. We are able to write:

$\small{\frac{1}{u(u+1)}\,=\,\frac{A_1}{u}~+~\frac{A_2}{u+1}}$

Where A₁ and A₂ are real numbers.

5. To find A₁ and A₂, we make denominators same on both sides:

$\small{\frac{1}{u(u+1)}\,=\,\frac{A_1}{u}~+~\frac{A_2}{u+1}\,=\,\frac{A_1(u+1)~+~A_2 u}{u(u+1)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:

$\small{1~=~A_1(u+1)~+~A_2 u}$

7. After equating the numerators, we can use suitable substitution.

♦ Put u = −1. We get: A₂ = −1

♦ Put u = 0. We get: A₁ = 1

8. Now the result in (4) becomes:

$\small{\frac{1}{u(u+1)}\,=\,\frac{1}{u}~-~\frac{1}{u+1}}$

9. So the integration becomes easy. We get:

$\small{\frac{1}{n}\left[\log \left|u \right| - \log \left|u+1 \right|~+~\rm{C_1} \right]}$

• The reader may write all the steps involved in the integration process.

10. Substituting for u, we get:

$\small{\frac{1}{n} \left[\log \left|x^n \right| - \log \left|x^n + 1 \right| \right]~+~\frac{\rm{C_1}}{n}}$

$\small{~=~\frac{1}{n}\log \left| \frac{x^n}{x^n + 1} \right|~+~\rm{C}}$

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The link below gives a few more miscellaneous examples:

Exercise 23.5

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We have completed a discussion on integration by partial fraction decomposition. In the next section, we will see integration by separation.

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