Question

How do you find the integral of `int sqrt(arctanx) / (x^2+1) dx` from 1 to infinity?

Answer 1

The integral has a value of `0.848` approximately.

Explanation:

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

`I = int_(pi/4)^oo sqrt(u) du`

`I = [2/3u^(3/2)]_(pi/4)^oo`

`I = lim_(t-> oo) [2/3(arctanx)^(3/2))]_1^oo`

Recall that `lim_(x->oo) arctanx = pi/2`.

`I = 2/3(pi/2)^(3/2) - 2/3(pi/4)^(3/2)`

`I~~ 0.848`

Hopefully this helps!