<s>[INST] Integrating $\int { \arctan{t} \over \sqrt{t} } $ 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.
 [/INST] Another simplification can be done for the term
$$ \int \frac{y^2}{1+y^4} dy$$
As
Divde by $y^2$
$$\int \frac{dy}{y^2 + \frac{1}{y^2}}$$
Following by
$$ \frac{1}{2}\int \frac{2+\frac{1}{y^2} - \frac{1}{y^2}}{y^2+\frac{1}{y^2}+2 -2}$$
Separating
$$\frac{1}{2}\left [\int \frac{1-\frac{1}{y^2}}{\left(y+\frac{1}{y}\right)^2 - (\sqrt{2})^2}dy + \frac{1+\frac{1}{y^2}}{\left(y-\frac{1}{y}\right)^2 + (\sqrt{2})^2}dy \right]$$
Let
$$I_1 = \frac{1}{2}\int \frac{1-\frac{1}{y^2}}{\left(y+\frac{1}{y}\right)^2 - (\sqrt{2})^2}dy$$
Now
$u =(y+ \frac{1}{y})$
And
$du = 1 - \frac{1}{y^2}$
$$ I_1 = \frac{1}{2}\int \frac{du}{u^2 - (\sqrt{2})^2}$$
Integrating, and
$$I_1 = \frac{1}{4\sqrt{2}} ln\left[\frac{u-\sqrt{2}}{u+ \sqrt{2}}\right] $$
Substituting the value of $u$
$$I_1 = \frac{1}{4\sqrt{2}} ln\left[\frac{y^2-\sqrt{2}y+1}{y^2+\sqrt{2}y+1}\right]. . .(1)$$
Now let
$$I_2 = \frac{1}{2} \int \frac{1+\frac{1}{y^2}}{\left(y-\frac{1}{y}\right)^2 + (\sqrt{2})^2}dy $$
Let,
$v = (y-\frac{1}{y})$
Also
$dv = 1+ \frac{1}{y^2}$
$$I_2 = \frac{1}{2}\int \frac{dv}{v^2 +(\sqrt{2})}$$
Integrating
$$ I_2 = \frac{1}{2}  \arctan \left(\frac{v}{\sqrt{2}}\right)$$
Substituting value of $v$
$$I_2 = \frac{1}{2\sqrt{2}} \arctan \left[\frac{y^2 -1}{\sqrt{2}y}\right] . . . (2)$$
From (1) and (2)
$$\int \frac{y^2}{1+y^4} dy = \frac{1}{4\sqrt{2}} ln\left[\frac{y^2-\sqrt{2}y+1}{y^2+\sqrt{2}y+1}\right] + \frac{1}{2\sqrt{2}} \arctan \left[\frac{y^2 -1}{\sqrt{2}y}\right] $$
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