Question

How to evaluate:`intx^5/(x^4 + 1)^2dx`?

Answer 1

`1/4tan^-1(x^2)-1/4 x^2/(x^4+1)+C`

Explanation:

Substitute `x^2=tan theta` in this, with `2xdx = sec^2theta d theta` to get

`int x^5/(x^4+1)^2 dx = 1/2 int x^4/(x^4+1)^2 2x dx`
`qquad = 1/2 int tan^2 theta/(tan^2 theta+1)^2 sec^2 theta d theta = 1/2 int (tan^2 theta sec^2 theta)/sec^4 theta d theta`
`qquad = 1/2int sin^2 theta d theta = 1/4 int (1-cos(2 theta)) d theta`
`qquad = 1/4 theta -1/8 sin(2 theta) = theta/4 -1/4 sin theta cos theta`
`qquad = theta/4 -1/4 tan theta/sec^2 theta = 1/4tan^-1(x^2)-1/4 x^2/(x^4+1)+C`