Question

How do you integrate `int (x+4) / [(x-1)(x^2+4)]` using partial fractions?

Answer 1

`int (x+4)/((x-1)(x^2+4)) = ln|x-1| - 1/2ln(x^2+4) +c`

Explanation:

The Partial fraction decomposition of the integrand will be of the form;

`(x+4)/((x-1)(x^2+4)) -= A/(x-1) + (Bx+C)/(x^2+4)`
`:. (x+4)/((x-1)(x^2+4)) = (A(x^2+4) + (Bx+C)(x-1))/((x-1)(x^2+4))`
`:. (x+4) -= A(x^2+4) + (Bx+C)(x-1)`

Put `x=1 => 5 = A(1+4) +0`
`:. 5A=5 => A=1`

Put `x=0 => 4=4A+(C)(-1)`
`:. C=4-4= 0`

Compare coefficients of `x^2 => 0 = A+B`
`:. B=-1`

Hence:
`(x+4)/((x-1)(x^2+4)) = 1/(x-1) -x/(x^2+4)`

And so,

`int (x+4)/((x-1)(x^2+4)) = int 1/(x-1) - int x/(x^2+4)`
`:. int (x+4)/((x-1)(x^2+4)) = int 1/(x-1) - 1/2int (2x)/(x^2+4)`
`:. int (x+4)/((x-1)(x^2+4)) = ln|x-1| - 1/2ln(x^2+4) +c`