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$y=arctanx$
$tany=x$

\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).

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    $\begingroup$ You need to change the integration measure as well, $dx=\sec^2{y} \, dy$. $\endgroup$ – Chappers Mar 29 '15 at 22:21
  • $\begingroup$ Lol, I can't believe I overlooked that. Thank you. $\endgroup$ – smokeypeat Mar 30 '15 at 0:59
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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.

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You may observe that $$ \left(-\frac{1}{(1+x^2)^{1/2}}\right)'=\frac {x}{{(1+x^2)}^{3/2}},\quad \left(\frac{x}{(1+x^2)^{1/2}}\right)'=\frac {1}{{(1+x^2)}^{3/2}}, $$ then just perform two integrations by parts as follows. $$ \begin{align} \int \frac { e^{\Large\arctan x}}{{(1+x^2)}^{3/2}}dx&=\frac {x\: e^{\Large\arctan x}}{{(1+x^2)}^{1/2}}-\int \frac { x\:e^{\Large\arctan x}}{{(1+x^2)}^{3/2}} \ dx\\\\ &=\frac {x\: e^{\Large\arctan x}}{{(1+x^2)}^{1/2}}-\left(-\frac { e^{\Large\arctan x}}{{(1+x^2)}^{1/2}}+\int \frac { e^{\Large\arctan x}}{{(1+x^2)}^{3/2}} \ dx\right)\\\\ &=\frac {(x+1)\: e^{\Large\arctan x}}{{(1+x^2)}^{1/2}}-\int \frac { e^{\Large\arctan x}}{{(1+x^2)}^{3/2}} \ dx \end{align}$$ giving easily $$\bbox[30px,border:1px solid #ff8000]{ \begin{align} \int \frac { e^{\Large\arctan x}}{{(1+x^2)}^{3/2}}dx=\frac {(x+1)\: e^{\Large\arctan x}}{2(1+x^2)^{1/2}}+C. \end{align}} $$

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