Question

How do you integrate `int x^3/(x^2+2)^3` by integration by parts method?

Answer 1

Integration by parts is: `intudv=uv-intvdu`
choose `u` so that `dv` can be integrated by variable substitution.

Explanation:

let `u=x^2 and dv=(x)/(x^2+2)^3dx`, because this will allow:

`v = intx/(x^2+2)^3dx`

to be integrated by letting `t = x^2+2`, then `dt = 2xdx` or `dt/2 = xdx`

This makes the integral become:

`v = 1/2intt^-3dt`

`v = -1/4t^-2`

Reversing the substitution:

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

and `du=2xdx`

`intx^3/(x^2+2)^3dx = (x^2)(-1/(4(x^2+2)^2)) - int-1/(x^2+2)^2(2x)dx`

`intx^3/(x^2+2)^3dx = -x^2/(4(x^2+2)^2) + 1/4int(2x)/(x^2+2)^2dx`

The last integral is the same sort of variable substitution:

`intx^3/(x^2+2)^3dx = -x^2/(4(x^2+2)^2) - 1/(4(x^2+2)) + C`