How do you integrate #int(6x^2+7x-6)/((x^2-4)(x+2))dx# using partial fractions?

2 Answers
Sep 22, 2017

Given: #int(6x^2+7x-6)/((x^2-4)(x+2))dx#

Decompose the integrand into partial fractions:

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

Multiply both sides by #(x-2)(x+2)^2#:

#6x^2+7x-6 = A(x+2)^2+B(x-2)+ C(x-2)(x+2)#

Find the value of A by letting #x = 2#:

#6(2)^2+7(2)-6 = A((2)+2)^2+B(2-2)+ C(2-2)(2+2)#

#32 = A(4)^2#

#A = 2#

Find the value of B by letting #x = -2#

#6(-2)^2+7(-2)-6 = 2(-2+2)^2+B(-2-2)+ C(-2-2)(-2+2)#

#4 = B(-2-2)#

#B = -1#

Find the value of C by letting #x = 0#

#6(0)^2+7(0)-6 = 2(0+2)^2-1(0-2)+ C(0-2)(0+2)#

#-6 = 8 + 2 -4C#

#C = 4#

Here is the decomposition:

#(6x^2+7x-6)/((x-2)(x+2)^2) = 2 1/(x-2)- 1/(x+2)^2+ 4 1/(x+2)#

Therefore, we can write the following equation:

#int(6x^2+7x-6)/((x^2-4)(x+2))dx = 2int 1/(x-2)dx - int 1/(x+2)^2 dx+ 4 int 1/(x+2)dx#

The integrals on the right are trival:

#int(6x^2+7x-6)/((x^2-4)(x+2))dx = 2ln|x-2| + 1/(x+2) + 4 ln|x+2| + C#

Sep 22, 2017

2 ln|x-2| +4 ln |x+2| + 1/(x+2) +C

Explanation:

First of all the denominator can be put in a different shape as #(x-2)(x+2)^2#. The partial fractions can thus be put as # A/(x-2) + (Bx)/(x+2)^2 +C/(x+2)#

The numerator would thus become #A(x+2)^2 +Bx(x-2) +C(x^2 -4)#

On simplification this adds up to #A(x^2+4x+4) +Bx^2 -2Bx +Cx^2 -4C# or, # (A+B+C)x^2 + (4A-2B)x +4A-4C.

Now comparing like terms with #6x^2 +7x -6#, it would be A+B+C=6, 4A-2B= 7 and4A-4C= -6. Solving these three equations, it would be A=2, B=#1/2# and C=#7/2#

The integral thus becomes #int 2/(x-2) dx +1/2 int x/(x+2)^2dx +7/2int1/(x+2) dx#

Or, #int 2/(x-2) dx + 1/2 int (x+2-2)/(x+2)^2 dx + 7/2 int 1/ (x+2) dx #

0r, #int 2/(x-2) dx +int 4/(x+2) dx - int 1/(x+2)^2 dx#

2 ln|x-2| +4 ln |x+2|+ 1/(x+2) +C Ans