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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23.14 - Solved Examples on Integration by Partial Fractions

In the previous section, we saw integration by partial fractions. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved Example 23.20
Find $\small{\int{\left[\frac{x^2 + 1}{x^2 - 5x + 6} \right]dx} }$
Solution:
1. The numerator is a polynomial of degree 2. The denominator is also a polynomial of degree 2.
2. So it is not a proper rational function. We must do long division. We get:
$\small{\frac{x^2 + 1}{x^2 - 5x + 6}\,=\,1~+~ \frac{5x - 5}{x^2 - 5x + 6}}$
• The reader may write all steps involved in the long division (or any other suitable method) process.

3. In the R.H.S, the first term can be easily integrated. But the second term must be subjected to partial fraction decomposition.
• First we factorize the denominator. We get:
x² − 5x + 6 = (x−2)(x−3)
• The reader may write all steps involved in the factorization process.   
   ♦ All factors are linear
   ♦ And all factors are distinct from one another.

4. So this is case I. We are able to write:
$\small{\frac{5x - 5}{x^2 - 5x + 6}\,=\,\frac{5x - 5}{(x-2)(x-3)}\,=\,\frac{A_1}{x-2}~+~\frac{A_2}{x-3}}$
Where A₁ and A₂ are real numbers.
5. To find A₁ and A₂, we make denominators same on both sides:
$\small{\frac{5x - 5}{(x-2)(x-3)}\,=\,\frac{A_1}{x-2}~+~\frac{A_2}{x-3}~=~\frac{A_1 (x-3)~+~A_2 (x-2)}{(x-2)(x-3)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:
$\small{5x - 5~=~A_1 (x-3)~+~A_2 (x-2)}$
7. After equating the numerators, we can use suitable substitution.
   ♦ Put x = 2. We get: 5 = A₁(−1). So A₁ = −5
   ♦ Put x = 3. We get: 10 = A₂(1). So A₂ = 10

8. Now the result in (2) becomes:
$\small{\frac{x^2 + 1}{x^2 - 5x + 6}\,=\,1~+~ \frac{5x - 5}{x^2 - 5x + 6}~=~1~+~\frac{(-5)}{x-2}~+~\frac{10}{x-3}}$

9. So the integration becomes easy. We get:
$\small{x - 5 \log \left|x-2 \right| + 10 \log \left|x-3 \right|~+~\rm{C}}$

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

Solved Example 23.21

Find $\small{\int{\left[\frac{1}{x^2 - 9} \right]dx}}$
Solution:
1. 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. We get:
x² − 9 = (x+3)(x−3)

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

4. So we are able to write:

$\small{\frac{1}{x^2 - 9}\,=\,\frac{1}{(x+3)(x-3)}\,=\,\frac{A_1}{x+3}~+~\frac{A_2}{x-3}}$

Where A₁ and A₂ are real numbers.

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

$\small{\frac{1}{(x+3)(x-3)}\,=\,\frac{A_1}{x+3}~+~\frac{A_2}{x-3}~=~\frac{A_1 (x-3)~+~A_2 (x+3)}{(x+3)(x-3)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:
$\small{1~=~A_1 (x-3)~+~A_2 (x+3)}$

7. After equating the numerators, we can use suitable substitution.
   ♦ Put x = 3. We get: 1 = A₂(6). So A₂ = 1/6  
   ♦ Put x = −3. We get: 1 = A₁(−6). So A₁ = −(1/6)

8. Now the result in (4) becomes:

$\small{\frac{1}{x^2 - 9}\,=\,\frac{1}{(x+3)(x-3)}\,=\,\frac{-1}{6(x+3)}~+~\frac{1}{6(x-3)}}$

9. So the integration becomes easy. We get:

$\small{\frac{1}{6}\log \left| \frac{x-3}{x+3}  \right|~+~\rm{C}}$

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

• Note that, in this solved example, we apply:
   ♦ Case I for the factors (x+3)(x−3)

Solved Example 23.22
Find $\small{\int{\left[\frac{3x - 1}{(x-1)(x-2)(x-3)} \right]dx}}$
Solution:
1. The numerator is a polynomial of degree 1. The denominator is a polynomial of degree 3.

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:

(x-1)(x-2)(x−3)

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

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

$\small{\frac{3x-1}{(x-1)(x-2)(x-3)}\,=\,\frac{A_1}{x-1}~+~\frac{A_2}{x-2}~+~\frac{A_3}{x-3}}$

Where A₁, A₂ and A3q are real numbers.

5. To find A₁, A₂ and A3q, we make denominators same on both sides:

$\small{\frac{3x-1}{(x-1)(x-2)(x-3)}\,=\,\frac{A_1 (x-2)(x-3)~+~A_2 (x-1)(x-3)~+~A_3 (x-1)(x-2)}{(x-1)(x-2)(x-3)}}$

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

$\small{3x-1~=~A_1 (x-2)(x-3)~+~A_2 (x-1)(x-3)~+~A_3 (x-1)(x-2)}$

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

   ♦ Put x = 2. We get: 5 = A₂(−1). So A₂ = −5
   
   ♦ Put x = 3. We get: 8 = A3q(2). So A3q = 4
   
   ♦ Put x = 1. We get: 2 = A₁(2). So A₁ = 1

8. Now the result in (4) becomes:

$\small{\frac{3x-1}{(x-1)(x-2)(x-3)}\,=\,\frac{1}{x-1}~+~\frac{-5}{x-2}~+~\frac{4}{x-3}}$

9. So the integration becomes easy. We get:

$\small{\log \left|x-1 \right| -5 \log \left|x-2 \right| + 4 \log \left|x-3 \right|~+~\rm{C}}$

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

• Note that, in this solved example, we apply:

♦ Case I for the factors (x−1)(x−2)(x−3)

Solved Example 23.23
Find $\small{\int{\left[\frac{x}{(x-1)(x-2)(x-3)} \right]dx}}$
Solution:
1. The numerator is a polynomial of degree 1. The denominator is a polynomial of degree 3.

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:

(x-1)(x-2)(x−3)

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

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

$\small{\frac{x}{(x-1)(x-2)(x-3)}\,=\,\frac{A_1}{x-1}~+~\frac{A_2}{x-2}~+~\frac{A_3}{x-3}}$

Where A₁, A₂ and A3q are real numbers.

5. To find A₁, A₂ and A3q, we make denominators same on both sides:

$\small{\frac{x}{(x-1)(x-2)(x-3)}\,=\,\frac{A_1 (x-2)(x-3)~+~A_2 (x-1)(x-3)~+~A_3 (x-1)(x-2)}{(x-1)(x-2)(x-3)}}$

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

$\small{x~=~A_1 (x-2)(x-3)~+~A_2 (x-1)(x-3)~+~A_3 (x-1)(x-2)}$

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

   ♦ Put x = 2. We get: 2 = A₂(−1). So A₂ = −2
   
   ♦ Put x = 3. We get: 3 = A3q(2). So A3q = 3/2
   
   ♦ Put x = 1. We get: 1 = A₁(2). So A₁ = 1/2

8. Now the result in (4) becomes:

$\small{\frac{x}{(x-1)(x-2)(x-3)}\,=\,\frac{1}{2(x-1)}~+~\frac{-2}{x-2}~+~\frac{3}{2(x-3)}}$

9. So the integration becomes easy. We get:

$\small{\frac{1}{2} \log \left|x-1 \right| -2 \log \left|x-2 \right| + \frac{3}{2} \log \left|x-3 \right|~+~\rm{C}}$

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

• Note that, in this solved example, we apply:

   ♦ Case I for the factors (x−1)(x−2)(x−3)

Solved Example 23.24
Find $\small{\int{\left[\frac{2x}{x^2 + 3x + 2} \right]dx}}$
Solution:
1. 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:
x² +3x + 2 = (x+1)(x+2)

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

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

$\small{\frac{2x}{x^2 + 3x + 2}\,=\,\frac{2x}{(x+1)(x+2)}\,=\,\frac{A_1}{x+1}~+~\frac{A_2}{x+2}}$

Where A₁ and A₂ are real numbers.

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

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

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

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

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

   ♦ Put x = −2. We get: −4 = A₂(−1). So A₂ = 4
   
   ♦ Put x = −1. We get: −2 = A₁(1). So A₁ = −2

8. Now the result in (4) becomes:

$\small{\frac{2x}{x^2 + 3x + 2}\,=\,\frac{2x}{(x+1)(x+2)}\,=\,\frac{-2}{x+1}~+~\frac{4}{x+2}}$

9. So the integration becomes easy. We get:

$\small{-2 \log \left|x+1  \right| + 4 \log \left|x+2  \right|~+~\rm{C}}$

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

• Note that, in this solved example, we apply:
   ♦ Case I for the factors (x+1)(x+2)

Solved Example 23.25

Find $\small{\int{\left[\frac{1-x^2}{x(1 - 2x)} \right]dx}}$
Solution:
1. The numerator is a polynomial of degree 2. The denominator is also a polynomial of degree 2.

2. So it is not a proper rational function. We must do long division first. We get:

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

3. The second term must be subjected to partial fraction decomposition.

• The denominator is already factorized. We have:

Denominator = 2x(1 − 2x)

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

4. So we are able to write:

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

Where A₁ and A₂ are real numbers.

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

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

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

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

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

   ♦ Put x = 0. We get: 2 = A₁
   
   ♦ Put x = 1/2. We get: 3/2 = A₂

8. Now the result in (4) becomes:

$\small{\frac{2-x}{2x(1 - 2x)}\,=\,\frac{2}{2x}~+~\frac{3}{2(1-2x)}}$

$\small{\implies \frac{2-x}{2x(1 - 2x)}\,=\,\frac{1}{x}~+~\frac{3}{2(1-2x)}}$

9. So the result in (5) becomes:

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

$\small{\,=\,\frac{1}{2}~+~\frac{1}{x}~+~\frac{3}{2(1-2x)}}$

10. Now the integration becomes easy. We get:

$\small{\frac{x}{2}~+~ \log \left|x \right|~-~\frac{3}{4} \log \left|1-2x  \right|~+~\rm{C}}$

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

• Note that, in this solved example, we apply:

   ♦ Case I for the factors 2x(1-2x)

Solved Example 23.26
Find $\small{\int{\left[\frac{x}{(x^2 + 1)(x - 1)} \right]dx}}$
Solution:
1. The numerator is a polynomial of degree 1. The denominator is a polynomial of degree 3.

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

3. The denominator is already factorized:
(x²+1)(x−1)

   ♦ One factor is quadratic.
   
   ♦ That quadratic factor cannot be further factorized.
   
   ♦ All other factors are linear.

4. The quadratic factor (x²+1) cannot be further factorized. So this is case III. Then we are able to write:

$\small{\frac{x}{(x^2 + 1)(x - 1)}~=~\left[\frac{Ax + B}{x^2 + 1}\right]~+~\frac{A_1}{ x - 1}}$
Where A, B and A₁ are real numbers.

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

$\small{\frac{x}{(x^2 + 1)(x - 1)}~=~\frac{(Ax+B) (x-1)~+~A_1 (x^2+1)}{(x^2 +1)(x-1)}}$

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

$\small{x~=~(Ax+B) (x-1)~+~A_1 (x^2+1)}$

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

   ♦ Put x = 1. We get: 1 = 2A₁. So A₁ = 1/2
   
   ♦ Put x = 0. We get: 0 = −B + 1/2. So B = 1/2
   
   ♦ Put x = −1. We get: −1 = (−A + (1/2))(-2) + (1/2)2. So A = −(1/2)

8. Now the result in (4) becomes:

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

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

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

9. So the integration becomes easy. We get:

$\small{\frac{1}{2} \tan^{-1}x~-~\frac{1}{4} \log (x^2+1)~+~\frac{1}{2} \log \left|x-1 \right|~+~\rm{C}}$

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

Solved Example 23.27
Find $\small{\int{\left[\frac{2}{(x^2 + 1)(1 - x)} \right]dx}}$
Solution:
1. The numerator is a polynomial of degree zero. The denominator is a polynomial of degree 3.

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

3. The denominator is already factorized:
(x²+1)(1−x)

   ♦ One factor is quadratic.
   
   ♦ That quadratic factor cannot be further factorized.
   
   ♦ All other factors are linear.

4. The quadratic factor (x²+1) cannot be further factorized. So this is case III. Then we are able to write:

$\small{\frac{2}{(x^2 + 1)(1 - x)}~=~\left[\frac{Ax + B}{x^2 + 1}\right]~+~\frac{A_1}{ 1 - x}}$
Where A, B and A₁ are real numbers.

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

$\small{\frac{2}{(x^2 + 1)(1 - x)}~=~\frac{(Ax+B) (1-x)~+~A_1 (x^2+1)}{(x^2 +1)(1-x)}}$

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

$\small{2~=~(Ax+B) (1-x)~+~A_1 (x^2+1)}$

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

   ♦ Put x = 1. We get: 2 = 2A₁. So A₁ = 1
   
   ♦ Put x = 0. We get: 2 = B + 1. So B = 1
   
   ♦ Put x = −1. We get: 2 = (−A + 1)(2) + 2. So A = 1

8. Now the result in (4) becomes:

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

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

9. So the integration becomes easy. We get:

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

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

Solved Example 23.28
Find $\small{\int{\left[\frac{1}{x(x^2 + 1)} \right]dx}}$
Solution:
1. The numerator is a polynomial of degree zero. The denominator is a polynomial of degree 3.

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

3. The denominator is already factorized:
x(x²+1)

   ♦ One factor is quadratic.
   
   ♦ That quadratic factor cannot be further factorized.
   
   ♦ All other factors are linear.

4. The quadratic factor (x²+1) cannot be further factorized. So this is case III. Then we are able to write:

$\small{\frac{1}{x(x^2 + 1)}~=~\left[\frac{Ax + B}{x^2 + 1}\right]~+~\frac{A_1}{x}}$
Where A, B and A₁ are real numbers.

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

$\small{\frac{1}{(x^2 + 1)(x)}~=~\frac{(Ax+B) (x)~+~A_1 (x^2+1)}{(x^2 +1)(x)}}$

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

$\small{1~=~(Ax+B) (x)~+~A_1 (x^2+1)}$

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

   ♦ Put x = 0. We get: 1 = A₁.
   
   ♦ Put x = 1. We get: 1 = A+B + 2. So A+B = −1
   
   ♦ Put x = −1. We get: 1 = (A − B) + 2. So A−B = −1
   
Solving the last two equations,we get: A = −1 and B = 0.   

8. Now the result in (4) becomes:

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

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

9. So the integration becomes easy. We get:

$\small{\log \left|x \right|~-~\frac{1}{2} \log (x^2+1)~+~\rm{C}}$

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

Solved Example 23.29
Find $\small{\int{\left[\frac{x}{(x-1)(x-2)} \right]dx}}$
Solution:
1. 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. The denominator is already in the factorized form:
(x−1)(x−2)

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

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

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

Where A₁ and A₂ are real numbers.

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

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

6. Since denominators are same on both sides, we can equate the numerators. We get:
$\small{x~=~A_1 (x-2)~+~A_2 (x-1)}$

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

   ♦ Put x = 2. We get: 2 = A₂.
   
   ♦ Put x = 1. We get: 1 = A₁(−1). So A₁ = −1

8. Now the result in (4) becomes:

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

9. So the integration becomes easy. We get:

$\small{\log \left| \frac{(x-2)^2}{x-1}  \right|~+~\rm{C}}$

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

---

In the next section, we will see some examples which involves substitution. 

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23.13 - Integration by Partial Fractions

In the previous section, we completed a discussion on standard integrals. In this section, we will see integration by partial fractions.

This method can be described in 5 steps:
1. We have seen rational functions in our earlier discussions. They are of the form $\small{\frac{\rm{P}(x)}{\rm{Q}(x)}}$.
• Where,
   ♦ P(x) is a polynomial function
   ♦ Q(x) is also a polynomial function
   ♦ Q(x) is not equal to zero
• An example: $\small{\frac{x^4 – 3 x^3 + 2x +5}{3 x^2 + 5x + 7}}$
2. There are two types of rational functions.
(i) If the degree of P(x) is less than the degree of Q(x), then it is called a proper rational function.
(ii) If the degree of P(x) is greater than or equal to the degree of Q(x), then it is called an improper rational function.

3. If we are asked to integrate an improper rational function, then we must first reduce it. This can be explained in 2 steps:
(i) Suppose that, we are given the rational function $\small{\frac{\rm{P}(x)}{\rm{Q}(x)}}$.
• where:
Degree of P(x) is greater than degree of Q(x).
(ii) Then we can perform long division (or any other suitable method) and write:
$\small{\frac{\rm{P}(x)}{\rm{Q}(x)}~=~\rm{T}(x)\,+\,\frac{\rm{P_1}(x)}{\rm{Q}(x)}}$
• Where:
   ♦ T(x) is the quotient of long division
   ♦ P₁(x) is the remainder of long division.
• An example:
$\small{\frac{4x^3 + 8x^2 + 8x + 7}{2x^2 – x + 1}\,=\,2x+5 \,+\,\frac{11x + 2}{2x^2 – x + 1}}$
• Here,
   ♦ T(x) = 2x + 5. It is the quotient  
   ♦ P₁(x) = 11x + 2. It is the remainder
(We have seen long division method in our earlier classes. Detailed notes can be seen here)
• In the reduced form, the integration of T(x) is easy.
• But the integration of $\small{\frac{\rm{P_1}(x)}{\rm{Q}(x)}}$ involves some additional steps (4) and (5) below.  
4. Our aim is to find a method to calculate $\small{\int{\left[\frac{\rm{P_1(x)}}{\rm{Q}(x)} \right]dx}}$
• The method involves decomposing $\small{\frac{\rm{P_1(x)}}{\rm{Q}(x)}}$ into simple rational functions.
• After the decomposition, $\small{\frac{\rm{P_1(x)}}{\rm{Q}(x)}}$ will be the sum of those simple rational functions.
• The method of decomposing, is called partial fraction decomposition.
• As a result of the decomposition, we get two or more simple rational functions. They will be so simple that, their integration can be achieved by the techniques that we saw in previous sections.

5. So our next task is to learn how to carry out the partial fraction decomposition.
• The first step is the factorization of the denominator Q(x). When we do this factorization, three cases can arise.
Case I:
This can be written in 5 steps:
(i) Case I arises when:
   ♦ All factors are linear,
   ♦ And the factors are distinct from one another.
(ii) So we can write:
Q(x) = (a₁x + b₁)(a₂x + b₂)(a₃x + b₃) . . . (a_nx + b_n)

(iii) In such a situation, we will be able to write:
$\small{\frac{\rm{P_1(x)}}{\rm{Q}(x)}~=~\frac{A_1}{a_1 x + b_1}\,+\,\frac{A_2}{a_2 x + b_2} \,+\, \frac{A_3}{a_3 x + b_3}\,+\,.~.~.~\,+\,\frac{A_n}{a_n x + b_n}}$
Where A₁, A₂, A₃ etc., are real numbers.
(We will see the proof in higher classes)
(iv) We can find the real numbers A₁, A₂, A₃ etc., by any one of the two methods:
   ♦ Comparing the coefficients.
   ♦ Suitable substitution.

(v) Once we find those real numbers, the decomposition process is over. We have converted $\small{\frac{\rm{P_1(x)}}{\rm{Q}(x)}}$ into a sum of simple rational functions.
• Let us see an example:

Solved Example 23.17
Find $\small{\int{\left[\frac{1}{x^2 + 3x + 2} \right]dx} }$
Solution:
1. The numerator is a polynomial of degree zero. The denominator is a polynomial of degree 2.
2. So it is a proper rational function. There is no need to do long division. We can straight away start partial fraction decomposition.
3. First we factorize the denominator. We get:
x² + 3x + 2 = (x+1)(x+2)
• The reader may write all steps involved in the factorization process.   
   ♦ All factors are linear
   ♦ And all factors are distinct from one another.
4. So we are able to write:
$\small{\frac{1}{x^2 + 3x + 2}\,=\,\frac{1}{(x+1)(x+2)}\,=\,\frac{A_1}{x+1}~+~\frac{A_2}{x+2}}$
Where A₁ and A₂ are real numbers.
5. To find A₁ and A₂, we make denominators same on both sides:
$\small{\frac{1}{(x+1)(x+2)}\,=\,\frac{A_1}{x+1}~+~\frac{A_2}{x+2}}$
⇒ $\small{\frac{1}{(x+1)(x+2)}\,=\,\frac{A_1(x+2)~+~A_2(x+1)}{(x+1)(x+2)}}$
6. Since denominators are same on both sides, we can equate the numerators. We get:
1 = A₁(x+2) + A₂(x+1)
⇒ 1 = A₁x + 2A₁ + A₂x + A₂
⇒ 0x + 1 = (A₁ + A₂)x + (2A₁ + A₂)
7. Comparing the coefficients in the above step, we get two equations:
(i) A₁ + A₂ = 0
(ii) 2A₁ + A₂ = 1
• Solving the two equations, we get:
A₁ = 1 and A₂ = −1
8. Alternatively, after equating the numerators, we can use suitable substitution.
We have: 1 = A₁(x+2) + A₂(x+1)
   ♦ Put x = −2. We get: 1 = A₂(−1) ⇒ A₂ = −1
   ♦ Put x = −1. We get: 1 = A₁(1) ⇒ A₁ = 1
9. Now the result in (4) becomes:
$\small{\frac{1}{x^2 + 3x + 2}\,=\,\frac{1}{(x+1)(x+2)}\,=\,\frac{1}{x+1}~+~\frac{-1}{x+2}}$
10. So the integration becomes easy. We get:
$\small{\int{\left[\frac{1}{x^2 + 3x + 2} \right]dx}\,=\,\int{\left[\frac{1}{x+1} \right]dx}~+~\int{\left[\frac{-1}{x+2} \right]dx}}$

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

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

Case II:
This can be written in 5 steps:
(i) Case II arises when:
   ♦ All factors are linear,
   ♦ And the factors are not distinct from one another.
(ii) So we can write:
Q(x) = (ax + b)ⁿ (a₁x + b₁)(a₂x + b₂)(a₃x + b₃) . . . (a_nx + b_n)
• Here, the factor (ax + b) occurs n times. So we cannot say that all factors are distinct from one another.

 
(iii) In such a situation, we will be able to write:
$\small{\frac{\rm{P_1(x)}}{\rm{Q}(x)}~=~\left[\frac{A_1}{a x + b}\,+\,\frac{A_2}{(a x + b)^2} \,+\, \frac{A_3}{(a x + b)^3}\,+\,.~.~.~\,+\,\frac{A_n}{(a x + b)^n}\right]}$

$\small{~+~\frac{B_1}{a_1 x + b_1}\,+\,\frac{B_2}{a_2 x + b_2} \,+\, \frac{B_3}{a_3 x + b_3}\,+\,.~.~.~\,+\,\frac{B_n}{a_n x + b_n}}$
• Where A₁, A₂, A₃ etc., and B₁, B₂, B₃ etc., are real numbers.
(We will see the proof in higher classes)

• Note that, for the repeating factor (ax+b), we provide a separate "sum" shown inside square brackets. The reader may observe it carefully and get a good feel of the pattern.

• For each repeating factor, we must provide it's own separate "sum" shown inside the square brackets.

(iv) As in case I, we can find the real numbers A₁, A₂, A₃ etc., and B₁, B₂, B₃ etc., by any one of the two methods:
   ♦ Comparing the coefficients.
   ♦ Suitable substitution.
(v) Once we find those real numbers, the decomposition process is over. We have converted $\small{\frac{\rm{P_1(x)}}{\rm{Q}(x)}}$ into a sum of simple rational functions.
• Let us see an example:

Solved Example 23.18
Find $\small{\int{\left[\frac{3x-2}{x^3 + 5x^2 + 7x + 3} \right]dx} }$
Solution:
1. The numerator is a polynomial of degree one. The denominator is a polynomial of degree 3.
2. So it is a proper rational function. There is no need to do long division. We can straight away start partial fraction decomposition.
3. First we factorize the denominator. We get:
x³ + 5x² + 7x + 3 = (x+1)²(x+3)
• The reader may write all steps involved in the factorization process.   
   ♦ All factors are linear
   ♦ And the factors are not distinct from one another.

4. The factor (x+1) occurs two times. So we are able to write:
$\small{\frac{3x-2}{(x+1)^2 (x+3)}~=~\left[\frac{A_1}{x + 1}\,+\,\frac{A_2}{(x + 1)^2}\right]~+~\frac{B_1}{x +3}}$
Where A₁,  A₂ and B₁ are real numbers.
5. To find A₁,  A₂ and B₁, we make denominators same on both sides:
$\small{\frac{3x-2}{(x+1)^2 (x+3)}~=~\frac{A_1 (x + 1) (x +3)~+~A_2(x+3)~+~B_1(x+1)^2}{(x + 1)^3(x+3)}}$
6. Since denominators are same on both sides, we can equate the numerators. We get:
$\small{3x-2\,=\,A_1 (x + 1) (x +3)~+~A_2(x+3)~+~B_1(x+1)^2}$

7. After equating the numerators, we can use suitable substitution.
   ♦ Put x = −1. We get: $\small{-3-2\,=\,A_2(-1+3)}$
So A₂ = −(5/2)
   ♦ Put x = −3. We get: $\small{-9-2\,=\,B_1(-2)^2}$
So B₂ = −(11/4)
   ♦ Put x = 0. We get: $\small{-2\,=\,3 A_1~+~3 A_2~+~B_1}$

⇒ $\small{-2\,=\,3 A_1~+~3 \left(\frac{-5}{2} \right)~+~\left(\frac{-11}{4} \right)}$

⇒ $\small{-2\,+\,\frac{15}{2}\,+\,\frac{11}{4}\,=\,3 A_1}$

⇒ $\small{A_1\,=\,\frac{11}{4}}$

8. Now the result in (4) becomes:
$\small{\frac{3x-2}{(x+1)^2 (x+3)}~=~\frac{11}{4(x + 1)}\,-\,\frac{5}{2(x + 1)^2}~-~\frac{11}{4(x +3)}}$

9. So the integration becomes easy. Integration of R.H.S will give:
$\small{\frac{11}{4} \log \left| \frac{x+1}{x+3}  \right|~+~\frac{5}{2(x+1)}~+~C}$

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

• Note that, in this solved example, we apply:
   ♦ Case II for the factor (x+2)²  
   ♦ Case I for the factor (x+3)  

Case III:
This can be written in 5 steps:
(i) Case III arises when:
   ♦ One or more factors are quadratic.
   ♦ Those quadratic factors cannot be further factorized.
   ♦ All other factors are linear.
(ii) So we can write:
Q(x) = (ax² + bx + c) (a₁x + b₁)(a₂x + b₂)(a₃x + b₃) . . . (a_nx + b_n)
 
(iii) In such a situation, we will be able to write:
$\small{\frac{\rm{P_1(x)}}{\rm{Q}(x)}~=~\left[\frac{Ax + B}{a x^2 + bx + c}\right]~+~\frac{A_1}{a_1 x + b_1}\,+\,\frac{A_2}{a_2 x + b_2} \,+\, \frac{A_3}{a_3 x + b_3}\,+\,.~.~.~\,+\,\frac{A_n}{a_n x + b_n}}$
Where A, B and A₁, A₂, A₃ etc., are real numbers.
(We will see the proof in higher classes)

• Note that, for the quadratic factor (ax² +bx + c), we provide a separate term shown inside square brackets. The reader may observe it carefully and get a good feel of the pattern.

• For each quadratic factor, we must provide it's own separate term shown inside the square brackets.

(iv) As in cases I and II, we can find the real numbers A, B and A₁, A₂, A₃ etc., by any one of the two methods:
   ♦ Comparing the coefficients.
   ♦ Suitable substitution.
(v) Once we find those real numbers, the decomposition process is over. We have converted $\small{\frac{\rm{P_1(x)}}{\rm{Q}(x)}}$ into a sum of simple rational functions.
• Let us see an example:

Solved Example 23.19
Find $\small{\int{\left[\frac{x^2 + x + 1}{x^3 + 2x^2 + x + 2} \right]dx} }$
Solution:
1. The numerator is a polynomial of degree 2. The denominator is a polynomial of degree 3.
2. So it is a proper rational function. There is no need to do long division. We can straight away start partial fraction decomposition.

3. First we factorize the denominator. We get:
x³ + 2x² + x + 2 = (x+2)(x²+1)
• The reader may write all steps involved in the factorization process.   
   ♦ One or more factors are quadratic.
   ♦ Those quadratic factors cannot be further factorized.
   ♦ All other factors are linear.
4. The quadratic factor (x²+1) cannot be further factorized. So we are able to write:
$\small{\frac{x^2 + x + 1}{(x+2)(x^2 + 1)}~=~\left[\frac{Ax + B}{x^2 + 1}\right]~+~\frac{A_1}{ x + 2}}$
Where A, B and A₁ are real numbers.
5. To find the real numbers, we make denominators same on both sides:
$\small{\frac{x^2 + x + 1}{(x+2)(x^2 + 1)}~=~\frac{(Ax + B)(x+2)~+~A_1(x^2 + 1)}{(x+2)(x^2 + 1)}}$
6. Since denominators are same on both sides, we can equate the numerators. We get:
$\small{x^2 + x + 1~=~(Ax + B)(x+2)~+~A_1(x^2 + 1)}$

7. After equating the numerators, we can use suitable substitution.
   ♦ Put x = −2. We get:
$\small{4 - 2 + 1~=~A_1(5)~~\implies~~A_1 = \frac{3}{5}}$
   ♦ Put x = 0. We get:
$\small{ 1~=~2B~+~A_1 \implies~~1 = 2B + \frac{3}{5}~~\implies B = \frac{1}{5} }$
   ♦ Put x = 1, and all known values. We get:
$\small{1^2 + 1 + 1~=~\left(A(1) + \frac{1}{5} \right)(1+2)~+~\frac{3}{5}(1^2 + 1)}$

$\small{\implies 3~=~3A + \frac{3}{5} + \frac{6}{5}\implies A = \frac{2}{5}}$

8. Now the result in (4) becomes:
$\small{\frac{x^2 + x + 1}{(x+2)(x^2 + 1)}~=~\left[\frac{(2/5)x + 1/5}{x^2 + 1}\right]~+~\frac{3/5}{ x + 2}}$

$\small{\implies \frac{x^2 + x + 1}{(x+2)(x^2 + 1)}~=~\frac{2x+1}{5(x^2 + 1)}~+~\frac{3}{ 5(x + 2)}}$

9. So the integration becomes easy. Integration of R.H.S can be done using known techniques. We get:
$\small{\frac{3}{5} \log \left|x+2 \right|~+~\frac{1}{5} \log \left|x^2 + 1 \right|~+~\frac{1}{5} \tan^{-1}x~+~\rm{C}}$

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

• Note that, in this solved example, we apply:
   ♦ Case III for the factor (x²+1)  
   ♦ Case I for the factor (x+2)

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We have seen all the three possible cases in partial fraction decomposition. In the next section, we will see a few more solved examples.

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