Question

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

Answer 1

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`

Answer 2

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