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# Integration by Partial Fractions – Integrals

• Difficulty Level : Medium
• Last Updated : 03 Jan, 2021

If f(x) and g(x) are polynomial functions such function. that g(x) â‰  0 then f(x)/g(x) is called Rational Functions. If degree f(x) < degree g(x) then f(x)/g(x) is called a proper rational function. If degree f(x) < degree g(x) then f(x)/g(x) is called an improper rational function. If f(x)/g(x) is an improper rational function then by dividing f(x) by g(x), we can express f(x)/g(x) as the sum of a polynomial and a proper rational function.

### Partial Fractions

Any proper rational function p(x)/q(x) can be expressed as the sum of rational functions, each having the simplest factor q(x), each such fraction is known as a partial function and the process of obtaining them is called the decomposition or resolving of the given function into partial fractions.

### Integration by Partial Fractions

For example lets say we want to evaluate âˆ«[p(x)/q(x)] dx where p(x)/q(x) is in a proper rational fraction. In cases like these, we can write the integrand as in a form of the sum of simpler rational functions by using partial fraction decomposition after that integration can be carried out easily. Here The values of A, B, C, etc. can be obtained as per the question.

### Examples

Example 1: Evaluate âˆ«(x – 1)/(x + 1)(x – 2) dx?

Solution:

Let (x – 1)/(x + 1)(x – 2) = A/(x + 1) + B/(x – 2)

Then, (x – 1) = A(x – 2) + B(x + 1) ………………(i)

Putting x = -1 in (i), we get A = 2/3

Putting x = 2 in (i), we get B = 1/3

Therefore,

(x – 1)/(x + 1)(x – 2) = 2/3(x + 1) + 1/3(x – 2)

=> âˆ«(x – 1)/(x + 1)(x – 2) = 2/3âˆ«dx/(x + 1) + 1/3âˆ«dx/(x – 2)

= 2/3log | x + 1 | + 1/3 log | x – 2 | + C

Example 2: Evaluate âˆ«dx/x{6(log x)2 + 7log x + 2}?

Solution:

Putting log x = t and 1/x dx = dt, we get

I = âˆ«dx/x{6(log x)2 + 7log x + 2} = âˆ«dt/(6t2 + 7t + 2) = âˆ«dt/(2t + 1)(3t + 2)

Let 1/(2t + 1)(3t + 2) = A/(2t + 1) + B/(3t + 2)

Then, 1 â‰¡ A(3t + 2) + B(2t + 1)  …………………….(i)

Putting t = -1/2 in (i), we get A = 2

Putting t = -2/3 in (i), we get B = -3

Therefore, 1/(2t + 1)(3t + 2) = 2/(2t + 1) + (-3)/(3t + 2)

=> I = âˆ«dt/(2t + 1)(3t + 2)

= âˆ«2dt/(2t + 1) – âˆ«3dt/(3t – 2)

= log | 2t + 1 | – log | 3t + 2 |

= log | 2t + 1 |/log | 3t + 2 | + C

= log | 2 log x + 1 | / log | 3 log x + 2 | + C

Example 3: Evaluate âˆ«dx/(x3 + x2 + x + 1)?

Solution:

We have 1/(x3 + x2 + x + 1) = 1/x2(x + 1) + (x + 1) = 1/(x + 1)(x2 + 1)

Let 1/(x + 1)(x2 + 1) = A/(x + 1) + Bx + C/(x2 + 1)  ……………………(i)

=> 1 â‰¡ A(x2 + 1) + (Bx + C) (x + 1)

Putting x = -1 on both sides of (i), we get A = 1/2.

Comparing coefficients of x2 on the both sides of (i), we get

A + B = 0 => B = -A = -1/2

Comparing coefficients of x on the both sides of (i), we get

B + C = 0 => C = -B = 1/2

Therefore, 1/(x + 1) (x2 + 1) = 1/2(x + 1) + (-1/2x + 1/2)/(x2 + 1)

Therefore, âˆ«1/(x + 1) (x2 + 1) = âˆ«dx/(x + 1) (x2 + 1)

= 1/2âˆ«dx/(x + 1) – 1/2âˆ«x/(x2 + 1)dx + 1/2âˆ«dx/(x2 + 1)

= 1/2âˆ«dx/(x + 1) – 1/4âˆ«2x/(x2 + 1)dx + 1/2âˆ«dx/(x2 + 1)

= 1/2 log | x + 1 | – 1/4 log | x2 + 1 | + 1/2 tan-1x + C

Example 4: Evaluate âˆ«x2/(x2 + 2)(x2 + 3)dx?

Solution:

Let x2/(x2 + 2) (x2 + 3) = y/(y + 2)(y + 3) where x2 = y.

Let y/(y + 2) (y + 3) = A/(y + 2) + B/(y + 3)

=> y â‰¡ A(y + 3) + B/(y + 2)   ………………(i)

Putting y = -2 on the both sides of (i), we get A = -2.

Putting y = -3 on the both sides of (i), we get B = 3.

Therefore, y/(y + 2) (y + 3) = -2/(y + 2) + 3/(y + 3)

=> x2/(x2 + 2) (x2 + 3) = -2/(x2 + 2) + 3/(x2 + 3)

=> âˆ«x2/(x2 + 2) (x2 + 3) = -2âˆ«dx/(x2 + 2) + 3âˆ«dx/(x2 + 3)

= -2/âˆš2tan-1(x/âˆš2) + 3/âˆš3tan-1(x/âˆš3) + C

= -âˆš2tan-1(x/âˆš2) + âˆš3tan-1(x/âˆš3) + C

Example 5: Evaluate âˆ«dx/x(x4 + 1)?

Solution:

We have

I = âˆ«dx/x(x4 + 1) = âˆ«x3/x4 (x4 + 1) dx [multiplying numerator and denominator by x3].

Putting x4 = t and 4x3dx = dt, we get

I = 1/4âˆ«dt/t(t + 1)

= 1/4âˆ«{1/t – 1/(t + 1)}dt   [by partial fraction]

= 1/4âˆ«1/t dt – 1/4âˆ«1/(t + 1)dt

= 1/4 log | t | – 1/4 log | t + 1 | + C

= 1/4 log | x4 | – 1/4 log | x4 + 1 | + C

= (1/4 * 4) log | x | – 1/4 log | x4 + 1 | + C

= log | x | – 1/4 log | x4 + 1 | + C

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