Integration Using Partial Fractions





In this page integration using partial fractions we are going to see how to integrate linear fractions using partial fractions.

Example 1:

Integrate (4x-3)/(x² + 3x + 8) with respect to x.

Solution:

(4x-3)/(x² + 3x + 8) dx

(4x-3)/(x² + 3x + 8) = { A [ d(x² + 3x + 8)/dx ] + B}/(x² + 3x + 8)

(4x-3) = A ( 2x + 3(1) + 0 ) ] + B

(4x-3)/(x² + 3x + 8) = [A ( 2x + 3 ) + B]/(x² + 3x + 8)

Equating the numerators

4x-3 = 2Ax + 6 + B

Equating the coefficient of x

4 = 2A                      -3 = 6 + B    

2A = 4                  6 + B = -3

 A = 4/2                     B = -3 - 6

 A = 2                        B = -9                

(4x-3)/(x² + 3x + 8) = [2 ( 2x + 3 ) - 9 ]/(x² + 3x + 8)

(4x-3)/(x² + 3x + 8) = [2 ( 2x + 3 )/(x² + 3x + 8)] - [9/(x² + 3x + 8)]

                   = [2 ( 2x + 3 )/(x² + 3x + 8)] dx - [9/(x² + 3x + 8)]dx

                   = 2 ( 2x + 3 )dx/(x² + 3x + 8)  - 9 1/(x² + 3x + 8)]dx

Now we have to integrate the two parts separately and then we have to combine.

             = 2 [(2x + 3)/(x² + 3x + 8)] dx  - 9 [1/(x² + 3x + 8)]dx -- (1)


Part 1:

 2 ∫[(2x + 3)/(x² + 3x + 8)] dx

Let u = x² + 3x + 8

    du = 2x + 3(1) + 0

    du = (2x + 3) dx

applying these values in the function

= 2 ∫ du/u

 = 2 log u

 = 2 log (x² + 3x + 8)


Part 2:

9 [1/(x² + 3x + 8)] dx

(x² + 3x + 8) = x² + (2/2) 3x + 8

                  = x² + 2 x x x (3/2)  + (3/2)² - (3/2)² + 8

                 = [x + (3/2)]² - 9/4 + 8

                 = [x + (3/2)]²  + (- 9 + 32)/4

                 = [x + (3/2)]²  + 23/4

                 = [x + (3/2)]²  + (√23)²/2²

                 = [x + (3/2)]²  + (√23/2)²]

                 = 9 [1/[(2x + 3)/2)]²  + (√23/2)²] dx

                 = 9 [1/[(2x + 3)²/4)]  + (√23)²/4] dx

                 = 9 x 4 [1/[(2x + 3)² + (√23)²] dx

                 = 9 x 4 [1/[(√23)² + (2x + 3)²] dx

∫ dx/( a² + ) = (1/a) tan⁻ ¹ (x/a) + c

a = √23    x = (2x + 3)

                  = 36 [(1/√23) tan⁻¹ ((2x + 3)/√23] + C

                  = 36 x (1/2) [(1/√23) tan⁻¹ ((2x + 3)/√23] + C

                  = 18 [(1/√23) tan⁻¹ [(2x + 3)/√23] + C

Applying these two answers in the first equation

= 2 log(x² + 3x + 8) + 18 [(1/√23) tan⁻¹ [(2x +3)/√23] + C

By practicing these kinds of problems we can easily do problems in integration using partial fractions

Related pages









Integration Using Partial Fractions to Integration