Question

How do you integrate `int (x^2)/(x+1)^3 dx` using partial fractions?

Answer 1

`int color(white)(.)x^2/(x+1)^3color(white)(.) dx = ln abs(x+1)+2/(x+1)-1/(2(x+1)^2)+C`

Explanation:

`int color(white)(.)x^2/(x+1)^3color(white)(.) dx = int color(white)(.)((x^2+2x+1)-(2x+2)+1)/(x+1)^3color(white)(.) dx`

`color(white)(int color(white)(.)x^2/(x+1)^3color(white)(.) dx) = int color(white)(.)((x+1)^2-2(x+1)+1)/(x+1)^3color(white)(.) dx`

`color(white)(int color(white)(.)x^2/(x+1)^3color(white)(.) dx) = int color(white)(.)1/(x+1)-2/(x+1)^2+1/(x+1)^3color(white)(.) dx`

`color(white)(int color(white)(.)x^2/(x+1)^3color(white)(.) dx) = ln abs(x+1)+2/(x+1)-1/(2(x+1)^2)+C`

Answer 2

The answer is `=-1/(2(x+1)^2)+(2)/(x+1)+ln(x+1)+C`

Explanation:

Let's do the decomposition in partial fractions
`x^2/(x+1)^3=A/(x+1)^3+B/(x+1)^2+C/(x+1)`
`=(A+B(x+1)+C(x+1)^2)/(x+1)^3`
We need to find A,B,C
So `x^2=A+B(x+1)+C(x+1)^2`
coefficients of `x^2` `=>` `C=1`
let `x=-1` `=>` `A=1`
Coefficients of x, `0=B+2C` `=>` `B=-2`
`:.x^2/(x+1)^3=1/(x+1)^3-2/(x+1)^2+1/(x+1)`
So `int(x^2dx)/(x+1)^3=intdx/(x+1)^3-int(2dx)/(x+1)^2+intdx/(x+1)`
We use `intx^ndx=x^(n+1)/(n+1)` and `intdx/x=lnx`
`int(x^2dx)/(x+1)^3=-1/(2(x+1)^2)+(2)/(x+1)+ln(x+1)+C`