Integrate: $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt$$

**My attempt:**

I first **substituted $y = \sqrt{t}$**

then $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt = 2\int { \arctan{y^2}}\,\mathrm dy$$

then I integrated **by parts**:

$u = \arctan{y^2}$, $v' = 1$
$$ = 2\left(y\arctan{y^2} - \int { y^2 \over 1+y^4}\,\mathrm dy\right)$$

but I am stuck, I cannot think of a way to solve this other integral.
I believe there is a more straightforward approach than the one I am using.

Thanks in advance.