How to integrate the function $\ f(x)=\arctan(\sqrt{\sqrt{x}-1}) $ How to integrate the function $\ f(x)=\arctan(\sqrt{\sqrt{x}-1}) $ 
I tried by $\ u=\arctan(\sqrt{\sqrt{x}-1}) $ and $\ dv=dx $, but wolfram alpha doesn't agree.
My solution is $\arctan(\sqrt{\sqrt{x}-1}) \cdot x -\frac{1}{2}\sqrt{x}-\frac{1}{2}\ln(\sqrt{x}-1)$

Integrate by parts:
$\int f \: \text{dg} = fg -\int g \: \text{df} $
where
$f = \arctan(\sqrt{\sqrt{x}-1})$
$\text{dg} = \text{dx}$
$\text{df} = \frac{1}{4x \sqrt{\sqrt{x}-1}} \text{dx}$
$g=x$
This gives
$$\int \arctan(\sqrt{\sqrt{x}-1})\: \text{dx} = x \arctan(\sqrt{\sqrt{x}-1})
- \int \frac{1}{4x \sqrt{\sqrt{x}-1}}\: \text{dx}$$
Focusing on the remaining integral, substitute $u = \sqrt{x}$.
$$
- \frac{1}{2} \int \frac{u}{\sqrt{u-1}} \: \text{du}$$
substitute $s = u-1$
$$
- \frac{1}{2} \int \sqrt{s} + \frac{1}{\sqrt{s}} \: \text{ds}$$
Now it should be straight forward to finish.
