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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