Question

Find `int \ x^5/(1+x^2)^4 \ dx`?

Answer 1

`int \ x^5/(1+x^2)^4 \ dx = - (3x^4+3x^2+1)/(6(1+x^2)^3) + C`

Explanation:

We want to find:

`I = int \ x^5/(1+x^2)^4 \ dx`

Let us perform a substitution:

Let `u=1+x^2 => (du)/dx=2x; \ \ \ x^2=u-1`

We can rewrite the integral and substitute as follows

`I = int \ 1/2 (x^2)^2/(1+x^2)^4 \ (2x) \ dx`
`\ \ = 1/2 \ int \ (u-1)^2/(u)^4 \ du`
`\ \ = 1/2 \ int \ (u^2-2u+1)/(u^4) \ du`
`\ \ = 1/2 \ int \ 1/u^2-2/u^3+1/(u^4) \ du`

Which we can now integrate to get:

`I = 1/2 (-1/u +1/u^2 - 1 /(3u^3) ) + C`
`\ \ = -1/2 (1/u -1/u^2 + 1 /(3u^3) ) + C`
`\ \ = -1/2 ( (3u^2-3u+1)/(3u^3) ) + C`
`\ \ = -1/6 ( (3u^2-3u+1)/(u^3) ) + C`

And by restoring the substitution we get:

`I = -1/6 ( (3(1+x^2)^2-3(1+x^2)+1)/((1+x^2)^3) ) + C`
`\ \ = -1/6 ( (3x^4+6x^2+3-3-3x^2+1)/((1+x^2)^3) ) + C`
`\ \ = -1/6 ( (3x^4+3x^2+1)/((1+x^2)^3) ) + C`
`\ \ = -1/6 ( (3x^4+3x^2+1)/((1+x^2)^3) ) + C`