23.12 - Some More Solved Examples on Standard Integrals

In the previous section, we saw some additional solved examples related to the standard integrals. In this section, we will see a few more solved examples.

Solved example 23.16
Find the following integrals:
$\small{(i)~\int{\left[\frac{dx}{\sqrt{7 - 6x - x^2}} \right]}~~~(ii)~\int{\left[\frac{4x + 1}{\sqrt{2x^2 + x - 3}} \right]dx}}$

$\small{(iii)~\int{\left[\frac{x+2}{\sqrt{x^2 - 1}} \right]dx}~~~(iv)~\int{\left[\frac{x+2}{\sqrt{x^2 + 2x+ 3}} \right]dx}}$
Solution:
Part (i):
1. We have the following data:
a = -1, b = -6 and c = 7

2. Based on this data, we can calculate two items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x +3}$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,-16}$

3. For this problem, we need to use formula VIII:
$\small{\int{\left[\frac{dx}{\sqrt{ax^2 + bx + c}} \right]~=~\frac{1}{\sqrt{a}} \int{\left[\frac{du}{\sqrt{u^2~\pm~k^2}}\right]}}}$

[Recall that in 2(1), we put u = x + b/(2a). So du = dx]

(i) So we want

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

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

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

(ii) Here we need to use formula V.

(iii) Let $\small{t = x+3}$

(iv) Then $\small{\frac{dt}{dx}=1~⇒~dx\,=\,dt }$

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

$\small{~=~\sin^{-1}\frac{ x+3}{4}+ \rm{C}}$

Part (ii): $\small{\int{\left[\frac{4x + 1}{\sqrt{2x^2 + x - 3}} \right]dx}}$

1. We have the following data:
a = 2, b = 1, c = -3, p = 4 and q = 1

2. Based on this data, we can calculate four items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x +\frac{1}{4}}$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,\frac{-25}{16}}$

(iii) $\small{A\,=\,\frac{p}{2a}\,=\,1}$

(iv) $\small{B\,=\,q\,-\,Ab\,=\,0}$

3. For this problem, we need to use formula X:
$\small{\int{\left[\frac{px + q}{\sqrt{ax^2 + bx + c}} \right]dx}~=~\int{\left[\frac{A(2ax + b)\,+\,B}{\sqrt{ax^2 + bx + c}} \right]dx}}$

4. So we need to calculate

$\small{\int{\left[\frac{1(4x \,+\, 1)\,+0}{\sqrt{2x^2 + x - 3}} \right]dx}~=~\int{\left[\frac{4x \,+\, 1}{\sqrt{2x^2 + x - 3}} \right]dx}}$

• The R.H.S has only one term

5. Calculation of the R.H.S:

(i) Let $\small{t=2x^2 + x - 3}$

(ii) Then $\small{\frac{dt}{dx}=4x + 1~⇒~(4x+1)dx\,=\,dt }$

(iii) $\small{\int{\left[\frac{1}{\sqrt{t}} \right]dt}= \frac{t^{1/2}}{1/2}\,+\,\rm{C}}$

$\small{~=~(2)\,t^{1/2}\,+\,\rm{C}~=~2\,\sqrt{2x^2 + x - 3 }\,+\,\rm{C}}$

Part (iii): $\small{\int{\left[\frac{x+2}{\sqrt{x^2 - 1}} \right]}}$

1. We have the following data:
a = 1, b = 0, c = -1, p = 1 and q = 2

2. Based on this data, we can calculate four items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x }$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,-1}$

(iii) $\small{A\,=\,\frac{p}{2a}\,=\,\frac{1}{2}}$

(iv) $\small{B\,=\,q\,-\,Ab\,=\,2}$

3. For this problem, we need to use formula X:
$\small{\int{\left[\frac{px + q}{\sqrt{ax^2 + bx + c}} \right]dx}~=~\int{\left[\frac{A(2ax + b)\,+\,B}{\sqrt{ax^2 + bx + c}} \right]dx}}$

4. So we need to calculate

$\small{\int{\left[\frac{\frac{1}{2}(2x \,+\, 0)\,+2}{\sqrt{x^2 - 1}} \right]dx}~=~\int{\left[\frac{x }{\sqrt{x^2 - 1}} \right]dx}~+~\int{\left[\frac{2}{\sqrt{x^2 - 1}} \right]dx}}$

• The R.H.S has two terms

5. Calculation of first term in the R.H.S of (4):

(i) Let $\small{t=x^2 - 1}$

(ii) Then $\small{\frac{dt}{dx}=2x + 0~⇒~(2x)dx\,=\,dt }$

(iii) $\small{\int{\left[\frac{1}{2\sqrt{t}} \right]dt}= \left(\frac{1}{2} \right)\frac{t^{1/2}}{1/2}\,+\,\rm{C}}$

$\small{~=~t^{1/2}\,+\,\rm{C}~=~\sqrt{x^2\,-\,1 }\,+\,\rm{C_1}}$

6. Calculation of second term in the R.H.S of (4):

(i) Here we need to use formula VIII:

$\small{\int{\left[\frac{dx}{\sqrt{ax^2 + bx + c}} \right]~=~\frac{1}{\sqrt{a}} \int{\left[\frac{du}{\sqrt{u^2~\pm~k^2}}\right]}}}$

[Recall that in 2(1), we put u = x + b/(2a). So du = dx]

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

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

• Here we need to use formula IV.

(iii) We get:

$\small{\int{\left[\frac{dx}{\sqrt{\left( x \right)^2~-~\left(1 \right)^2 }}\right]}~=~ \log\left|x~+~\sqrt{x^2 - 1} \right|~+~\rm{C_2}}$

7. Now we can write the R.H.S of (4):

$\small{\sqrt{x^2\,-\,1 }\,+\,\rm{C_1}~+~\log\left|x~+~\sqrt{x^2 - 1} \right|+ \rm{C_2}}$

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

Part (iv): $\small{\int{\left[\frac{x+2}{\sqrt{x^2 + 2x + 3}} \right]dx}}$

1. We have the following data:
a = 1, b = 2, c = 3, p = 1 and q = 2

2. Based on this data, we can calculate four items:

(i) $\small{u \,=\, x + \frac{b}{2a}\,=\,x+1 }$

(ii) $\small{\pm k^2\,=\,\frac{c}{a}\,-\, \frac{b^2}{4 a^2}\,=\,2}$

(iii) $\small{A\,=\,\frac{p}{2a}\,=\,\frac{1}{2}}$

(iv) $\small{B\,=\,q\,-\,Ab\,=\,1}$

3. For this problem, we need to use formula X:
$\small{\int{\left[\frac{px + q}{\sqrt{ax^2 + bx + c}} \right]dx}~=~\int{\left[\frac{A(2ax + b)\,+\,B}{\sqrt{ax^2 + bx + c}} \right]dx}}$

4. So we need to calculate

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

• The R.H.S has two terms

5. Calculation of first term in the R.H.S of (4):

(i) Let $\small{t=x^2 + 2x + 3}$

(ii) Then $\small{\frac{dt}{dx}=2x + 2~⇒~(2x+2)dx\,=\,dt }$

(iii) $\small{\int{\left[\frac{1}{2\sqrt{t}} \right]dt}= \left(\frac{1}{2} \right)\frac{t^{1/2}}{1/2}\,+\,\rm{C}}$

$\small{~=~t^{1/2}\,+\,\rm{C}~=~\sqrt{x^2 + 2x + 3 }\,+\,\rm{C_1}}$

6. Calculation of second term in the R.H.S of (4):

(i) Here we need to use formula VIII:

$\small{\int{\left[\frac{dx}{\sqrt{ax^2 + bx + c}} \right]~=~\frac{1}{\sqrt{a}} \int{\left[\frac{du}{\sqrt{u^2~\pm~k^2}}\right]}}}$

[Recall that in 2(1), we put u = x + b/(2a). So du = dx]

(ii) So we want $\small{\left(\frac{1}{\sqrt{1}} \right)  \int{\left[\frac{dx}{\sqrt{\left( x+1 \right)^2~+~\left(\sqrt{2} \right)^2}}\right]}}$

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

• Here we need to use formula VI.

(iii) We get:

$\small{\int{\left[\frac{dx}{\sqrt{\left( x+1 \right)^2~+~\left(\sqrt{2} \right)^2 }}\right]}~=~ \log\left|x+1~+~\sqrt{(x+1)^2 + 2} \right|~+~\rm{C_2}}$

$\small{~=~ \log\left|x+1~+~\sqrt{x^2 + 2x + 3} \right|~+~\rm{C_2}}$

7. Now we can write the R.H.S of (4):

$\small{\sqrt{x^2\,+\,2x + 3}\,+\,\rm{C_1}~+~\log\left|x+1~+~\sqrt{x^2 + 2x + 3} \right|+ \rm{C_2}}$

$\small{~=~\sqrt{x^2\,+\,2x + 3}~+~\log\left|x+1~+~\sqrt{x^2 + 2x + 3} \right|\,+\, \rm{C}}$

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The link below gives a few more miscellaneous examples:

Exercise 23.4

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In the next section, we will see integration by partial fractions.

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