calculus - 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).

Answer

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
$$\frac{1}{2}e^y(\cos{y}+\sin{y})=\frac{1}{2}e^y \cos{y}(1+\tan{y}) = \frac{e^{\arctan{x}}}{2\sqrt{1+x^2}}(1+x)$$

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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.