**Integration Using Partial Fractions Examples :**

Here we are going to see some example problems in integration using the concept of partial fractions.

**Example 1 :**

Integrate the following function with respect to x :

1/(x - 1) (x + 2)^{2}

**Solution :**

Decompose the given rational function into partial fractions.

1/(x - 1) (x + 2)^{2 } = A/(x - 1) + B/(x + 2) + C/(x + 2)^{2}

1 = A(x+2)^{2} + B(x-1)(x+2) + C(x-1)

Plug x = 1

1 = 9A

A = 1/9

Plug x = -2

1 = C(-3)

C = -1/3

Plug x = 0

1 = 4A-2B-C

1 = 4(1/9) - 2B - (-1/3)

1 = (4/9)- 2B + (1/3)

1 - (4/9) - (1/3) = -2B

-2B = (9 - 4 - 3)/9

-2B = 2/9

B = -1/9

Integrate.

**∫[**1/(x - 1) (x + 2)^{2}] dx

= (1/9)**∫(1**/(x-1))dx-(1/9)**∫**(1/(x+2)-(1/3)**∫**(1/(x+2)^{2 }dx

= (1/9)log(x-1) - (1/9)log (x+2)-(1/3)(1/(x+2)) + c

= (1/9) [log(x - 1) - log(x + 2)] - (1/3)(1/(x+2)) + c

= (1/9) [log(x - 1)/(x - 2)] - (1/3(x + 2)) + c

**Example 2 :**

Integrate the following function with respect to x :

(3x - 9)/(x - 1)(x + 2)(x^{2} + 1)

**Solution :**

Decompose the given rational function into partial fractions.

(3x - 9)/(x - 1)(x + 2)(x^{2} + 1)

= A/(x - 1) + B/(x + 2) + (Cx + D)/(x^{2} + 1)

3x - 9 = A(x+2)(x^{2}+1)+B(x-1)(x^{2}+1)+(Cx+D)(x-1)(x+2)

Plug x = 1

3 - 9 = A(6)

-6 = 6A ==> A = -1

Plug x = -2

-6 - 9 = B(-3)(5)

-15 = -15B ==> B = 1

Plug x = 0

- 9 = 2A - B - 2D

-9 = 2(-1) - 1 - 2D

-9 + 3 = -2D

-6 = -2D

D = 3

3x - 9 = A(x+2)(x^{2}+1)+B(x-1)(x^{2}+1)+(Cx+D)(x-1)(x+2)

Plug x = -1

- 12 = A(1)(2) + B(-2)(2) + (-C+D)(-2)

-12 = 2A - 4B + 2C - 2D

-12 = 2(-1) - 4(1) + 2C - 2(3)

-12 = -2 - 4 - 6 + 2C

2C = -12 + 12

C = 0

Integrate.

**∫[**(3x - 9)/(x - 1)(x + 2)(x^{2} + 1)] dx

= -**∫1**/(x - 1) dx + **∫**1/(x + 2) dx + **∫ 3**/(x^{2} + 1) dx

= - log (x - 1) + log (x + 2) + 3 tan^{-1}(x)

= log (x + 2)/(x - 1) + 3 tan^{-1}(x) + c

**Example 3 :**

Integrate the following function with respect to x :

x^{3}/(x - 1)(x - 2)

**Solution :**

x^{3}/(x - 1)(x - 2) = x^{3}/(x^{2} - 3x + 2)

In the given rational fraction, the highest exponent of x in numerator is greater than the highest exponent of x in denominator.

So, we can use long division to decompose the given rational function.

From the above long division, we have

x^{3}/(x - 1)(x - 2) = (x + 3) + (7x - 6)/(x^{2} - 3x + 2)

x^{3}/(x - 1)(x - 2) = (x + 3) + (7x - 6)/(x - 1)(x - 2) ----(1)

Decompose (7x - 6)/(x - 1)(x - 2) into partial fractions.

(7x - 6)/(x - 1)(x - 2) = A/(x - 1) + B(x - 2)

Simplify.

7x - 6 = A(x - 2) + B(x - 1)

Plug x = 1

7(1) - 6 = A(1 -2 ) + B(1 - 1)

1 = A(-1)

A = -1

Plug x = 2

7(2) - 6 = A(2 -2 ) + B(2 - 1)

8 = B(1)

B = 8

Therefore we have

(1)-----> x^{3}/(x - 1)(x - 2) = (x + 3) - 1/(x - 1) + 8/(x - 2)

Integrate.

**∫[**x^{3}/(x - 1)(x - 2)] dx

= **∫**(x + 3) dx -**∫1**/(x - 1) dx + 8**∫1**/(x - 2) dx

= x^{2}/2 + 3x - log(x - 1) + 8log(x - 2) + C

After having gone through the stuff given above, we hope that the students would have understood, how to integrate rational functions using partial fractions.

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