Question

How do you integrate `int (x^2 + 5x - 7) /( x^2 (x+ 1)^2)` using partial fractions?

Answer 1

The answer is `=7/x+19ln(|x|)+11/(x+1)-19ln(|x+1|)+C`

Explanation:

Let's perform the decomposition into partial fractions

`(x^2+5x-7)/(x^2(x+1)^2)=A/x^2+B/x+C/(x+1)^2+D/(x+1)`

`=(A(x+1)^2+Bx(x+1)^2+Cx^2+Dx^2(x+1))/((x^2(x+1)^2))`

As the denominators are the same, we can compare the numerators

`x^2+5x-7=A(x+1)^2+Bx(x+1)^2+Cx^2+Dx^2(x+1)`

Let `x=0`, `=>`, `-7=A`

Let `x=-1`, `=>`, `-11=C`

Coefficients of `x^3`, `=>`, `0=B+D`

Coefficients of `x`, `=>`, `5=2A+B`

`B=5-2A=5+14=19`

`D=-B=-19`

Therefore,

`(x^2+5x-7)/(x^2(x+1)^2)=-7/x^2+19/x-11/(x+1)^2-19/(x+1)`

So,

`int((x^2+5x-dx)/(x^2(x+1)^2)=-7intdx/x^2+19intdx/x-11intdx/(x+1)^2-19intdx/(x+1)`

`=7/x+19ln(|x|)+11/(x+1)-19ln(|x+1|)+C`