How to evaluate this integral? $\int \frac {x e^{\arctan(x)}}{{(1+x^2)}^{3/2}} \ dx$ I am working on a integral and I run out of ideas how to solve it. Does anyone has good some good idea? I tried various substitutions but it seems that I did not find the correct one. 
$$\int \frac {x e^{\displaystyle\arctan(x)}}{{(1+x^2)}^{3/2}} \ dx$$
 A: Using Mr. Mark Bennet's suggestion. Let $x=\tan y$ and $dx=\sec^2 y\,dy$, then
\begin{align}
\int \frac {x e^{\Large\arctan(x)}}{{(1+x^2)}^{\Large\frac{3}{2}}} \ dx&=\int \frac {\tan y\, e^{\Large\arctan(\tan y)}}{{(1+\tan^2y)}^{\Large\frac{3}{2}}} \sec^2 y\,dy\\
&=\int \frac {\frac{\sin y}{\cos y}\, e^{y}}{\sec^3 y} \sec^2 y\,dy\\
&=\int e^{y}\sin y\,dy\\
\end{align}
Using integration by parts, let $u=\sin y$, $du=\cos y\ dy$, $dv=e^{y}\, dy$, and $v=e^{y}$.
\begin{align}
\int e^{y}\sin y\,dy
&=e^{y}\sin y-\int e^{y}\cos y\,dy\\
\end{align}
Once again using integration by parts, let $u=\cos y$, $du=-\sin y\ dy$, $dv=e^{y}\, dy$, and $v=e^{y}$.
\begin{align}
\int e^{y}\sin y\,dy
&=e^{y}\sin y-\int e^{y}\cos y\,dy\\
&=e^{y}\sin y-e^{y}\cos y-\int e^{y}\sin y\,dy+C_1\\
\int e^{y}\sin y\,dy+\int e^{y}\sin y\,dy&=e^{y}(\sin y-\cos y)+C_1\\
2\int e^{y}\sin y\,dy&=e^{y}(\sin y-\cos y)+C_1\\
\int e^{y}\sin y\,dy&=\frac{e^{y}}{2}(\sin y-\cos y)+C_2\\
\end{align}
Since $\tan y=x$, then $\sin y=\Large\frac{x}{\sqrt{1+x^2}}$ and $\cos y=\Large\frac{1}{\sqrt{1+x^2}}$.
\begin{align}
\int \frac {x e^{\Large\arctan(x)}}{{(1+x^2)}^{\Large\frac{3}{2}}} \ dx
&=\int e^{y}\sin y\,dy\\
&=\frac{e^{y}}{2}(\sin y-\cos y)+C_2\\
&=\frac{e^{\Large\arctan(x)}}{2}\frac{(x-1)}{\sqrt{1+x^2}}+C_2\\
&=\frac{(x-1)}{2\sqrt{x^2+1}}e^{\Large\arctan(x)}+C_2\\
\end{align}
Credit answer to Mr. Mark Bennet. \(^_^)/
A: Try $\arctan x = t$. Then use trigonometry to simplify and finally integrate by parts.
$$\mathrm{d}x/(1+x^2)=\mathrm{d}t$$
$$\int\!\frac{\tan {t} e^t}{\sec t}\mathrm{d}t=\int\! e^t\sin t \,\mathrm{d}t$$
Where I have used $\sqrt{1+\tan^2t}=...$
