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.
 A: Elaborating on the comment, the key is to note that the denominator can indeed be further factored
$$\begin{align*}y^4+1 & =y^4+2y^2+1-2y^2\\ & =\left(y^2+1\right)^2-2y^2\\ & =\left(y^2+y\sqrt2+1\right)\left(y^2-y\sqrt2+1\right)\end{align*}$$
Therefore, we can use standard partial fraction decomposition to arrive at
$$\frac {y^2}{y^4+1}=\frac {Ay+B}{y^2+y\sqrt2+1}+\frac {Cy+D}{y^2-y\sqrt2+1}=\frac 1{2\sqrt{2}}\left(\frac {y}{y^2-y\sqrt{2}+1}-\frac {y}{y^2+y\sqrt{2}+1}\right)$$

The algebra is a bit messy for the partial fraction decomposition. One cool method you can use to find the coefficients is, starting off with$$\frac {y^2}{(y^2-y\sqrt{2}+1)(y^2+y\sqrt2+1)}=\frac {Ay+B}{y^2+y\sqrt2+1}+\frac {Cy+D}{y^2-y\sqrt2+1}$$Equate one of the factors to zero, say $$y^2-y\sqrt{2}+1=0\qquad\implies\qquad y^2=y\sqrt{2}-1$$Now, covering up the $y^2-y\sqrt{2}+1$ factor in the denominator on the left-hand side, take the limit as $y^2$ tends towards $y\sqrt{2}-1$; the resulting expression will be your $Cy+D$ term. To put it in equation-form$$Cy+D=\lim\limits_{y^2\to y\sqrt{2}-1}\frac {y^2}{y^2+y\sqrt{2}+1}=\frac {y\sqrt2-1}{2y\sqrt2}=\frac {y}{2\sqrt{2}}$$To get to the last line, I expanded the fraction into its constituents and used the equation$$\frac 1y=\sqrt2-y$$Directly comparing coefficients, it's clear that we must have$$C=\frac 1{2\sqrt2}\qquad\qquad D=0$$Repeat the same procedure with $y^2\to -y\sqrt2-1$ to arrive at$$A=-\frac 1{2\sqrt2}\qquad\qquad B=0$$

The integral becomes
$$\int\frac {y^2}{y^4+1}\,\mathrm dy=\frac 1{2\sqrt{2}}\int\frac {y}{y^2-y\sqrt{2}+1}\,\mathrm dy-\frac 1{2\sqrt2}\int\frac {y}{y^2+y\sqrt{2}+1}\,\mathrm dy$$
The resulting integral can be evaluated by first observing that $(y^2\pm y\sqrt{2}+1)'=2y\pm\sqrt{2}$. Thus, adding and subtracting $1/\sqrt2$ from the numerator then gives
$$\begin{align*}\int\frac {y}{y^2\pm y\sqrt{2}+1}\,\mathrm dy & =\frac 12\int\frac {2y\pm\sqrt{2}\mp\sqrt{2}}{y^2\pm y\sqrt{2}+1}\,\mathrm dy\\ & =\frac 12\int\frac {2y\pm\sqrt{2}}{y^2\pm y\sqrt{2}+1}\,\mathrm dy\mp\frac 1{\sqrt{2}}\int\frac {\mathrm dy}{\left(y\pm\frac 1{\sqrt2}\right)^2+\frac 12}\\ & =\frac 12\log(y^2\pm y\sqrt{2}+1)\mp\arctan\left(y\sqrt{2}\pm 1\right)\end{align*}$$
Putting everything together, then we see that
$$\begin{align*}\int\frac {y^2}{y^4+1}\,\mathrm dy=\frac 1{4\sqrt2}\log\left(\frac {y^2-y\sqrt{2}+1}{y^2+y\sqrt{2}+1}\right) & +\frac 1{2\sqrt{2}}\arctan\left(y\sqrt{2}-1\right)\\ & +\frac 1{2\sqrt{2}}\arctan\left(y\sqrt{2}+1\right)+C\end{align*}$$
Thus, our integral becomes
$$\begin{align*}\int\arctan y^2\,\mathrm dy=y\arctan y^2-\frac 1{2\sqrt2}\log\left(\frac {y^2-y\sqrt2+1}{y^2+y\sqrt2+1}\right) & -\frac 1{\sqrt2}\arctan\left(y\sqrt2-1\right)\\ & -\frac 1{\sqrt2}\arctan\left(y\sqrt2+1\right)+C\end{align*}$$
Confirmed by Wolfram Alpha
A: 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)
$$\int \frac{y^2}{1+y^4} dy = \frac{1}{4\sqrt{2}} ln\left[\frac{y^2-\sqrt{2}y+1}{y^2+\sqrt{2}y+1}\right] + \frac{1}{2\sqrt{2}} \arctan \left[\frac{y^2 -1}{\sqrt{2}y}\right] $$
A: You can do it faster since
$$\frac {y^2}{1+y^4}=\frac {y^2}{(y^2+i)(y^2-i)}=\frac 1 2\Big[\frac 1{y^2+i} +\frac 1{y^2-i}\Big]$$ and
$$\int \frac {dy}{y^2+k\, i}=-\frac{1-i}{\sqrt{2k}}\tan ^{-1}\left(-\frac{1-i}{\sqrt{2k}}{y}\right)$$
