Skip to content
Related Articles

Related Articles

Improve Article

Integration by Partial Fractions – Integrals

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

If f(x) and g(x) are polynomial functions such function. that g(x) ≠ 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 ∫[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.

Factors in the denominator or denominators in rational functions

            Corresponding Partial Fractions             



(x – a)

 A/(x – a)

(x – b)

A/(x – b) + A/(x – b)2

(x – c)

A/(x – c) + B/(x – c)2 + C/(x – c)3

(ax2 + bx +c ) 

Ax + B/(ax2 + bx + c)



Examples

Example 1: Evaluate ∫(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)

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

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



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

Solution:  

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

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

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

Then, 1 ≡ 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 = ∫dt/(2t + 1)(3t + 2)

        = ∫2dt/(2t + 1) – ∫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 ∫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 ≡ 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, ∫1/(x + 1) (x2 + 1) = ∫dx/(x + 1) (x2 + 1)

                                              = 1/2∫dx/(x + 1) – 1/2∫x/(x2 + 1)dx + 1/2∫dx/(x2 + 1)

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

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

Example 4: Evaluate ∫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 ≡ 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)

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

                                     = -2/√2tan-1(x/√2) + 3/√3tan-1(x/√3) + C



                                     = -√2tan-1(x/√2) + √3tan-1(x/√3) + C

Example 5: Evaluate ∫dx/x(x4 + 1)?

Solution:

We have

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

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

I = 1/4∫dt/t(t + 1)

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

  = 1/4∫1/t dt – 1/4∫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

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.




My Personal Notes arrow_drop_up
Recommended Articles
Page :