Integrate $ \int \frac { e^{\arctan(x)}}{{(1+x^2)}^{\frac{3}{2}}} \ dx $ $y=arctanx$
$tany=x$
\begin{align}
\int \frac { e^{\Large\arctan(x)}}{{(1+x^2)}^{\Large\frac{3}{2}}} \ dx&=\int \frac {e^{\Large\arctan(\tan y)}}{{(1+\tan^2y)}^{\Large\frac{3}{2}}}dy\\
&=\int \frac {e^{y}}{\sec^3 y} dy\\
&= e^y \cos^3 y+ \int 3e^{y}\sin y\ \cos^2 y\ dy\\
\end{align}
Is this right so far or am I doing something wrong?
It's been quite a while since I've done integration with trig substitutions. Last time I did this integral I did not use trig subtitution and still got the correct answer, I can't find my solutions from then(over 2 years ago).

Your substitution gives
$$ I = \int \frac{e^y}{(1+\tan^2{y})^{3/2}} \sec^2{y} \, dy = \int e^y \cos{y} \, dy, $$
which we work out by integrating by parts a couple of times to be
$$ \frac{1}{2}e^y(\cos{y}+\sin{y})=\frac{1}{2}e^y \cos{y}(1+\tan{y}) = \frac{e^{\arctan{x}}}{2\sqrt{1+x^2}}(1+x) $$

You can also do this by parts without substitution: the derivative of $e^{\arctan{x}}$ is $e^{\arctan{x}}/(1+x^2)$, so you have
$$ I = \int \frac{1}{\sqrt{1+x^2}}\frac{e^{\arctan{x}}}{1+x^2} \, dx \\
= \frac{e^{\arctan{x}}}{\sqrt{1+x^2}} + \int \frac{x e^{\arctan{x}}}{(1+x^2)^{3/2}} \, dx $$
If you do this again, you get
$$ \int \frac{x}{\sqrt{1+x^2}} \frac{e^{\arctan{x}}}{1+x^2} \, dx = \frac{xe^{\arctan{x}}}{1+x^2} - \int \frac{e^{\arctan{x}}}{(1+x^2)^{3/2}} \left( 1+x^2-x^2 \right) \, dx, $$
and $I$ has reappeared on the right, so solving for $I$ gives
$$ I = \frac{1+x}{2\sqrt{1+x^2}}e^{\arctan{x}} $$
as before.
