Question

What is `int_0^sqrt(3) (x-(arctanx)^4)/(1+x^2)`?

Answer 1

The integral equals `0.441` most nearly.

Explanation:

We have:

`I = int_0^sqrt(3) x/(1 + x^2) - arctan^4x/(x^2 + 1)dx`

`I = int_0^sqrt(3) x/(1 + x^2) - int_0^sqrt(3) arctan^4x/(x^2 + 1) dx`

We now let `u = arctanx`. Then `du = 1/(1 + x^2) dx` and `dx = (1 + x^2)du`.

`I = int_0^sqrt(3) x/(1+ x^2) - int_0^(pi/3) u^4 du`

For the first integral, let `t = x^2 + 1`. Then `dt = 2x dx` and `dx = (dt)/(2x)`.

`I = int_1^4 x/(2tx) dt - int_0^(pi/3) u^4 du`

`I = [1/2ln|t|]_1^4 - [1/5u^5]_0^(pi/3)`

`I = 1/2ln4 - 1/5(pi/3)^5`

`I = ln2 - pi^5/1215 ~~0.441`

Hopefully this helps!