Evaluating $\int\frac{e^{\tan^{-1} x}}{(1+x^2)^{\frac32}}dx$ I have $$\int\dfrac{e^{\tan^{-1} x}}{(1+x^2)^{\tfrac32}}dx$$
I tried using Integral By Parts:
$$\int \dfrac{1}{\sqrt{1+x^2}}\times\dfrac{e^{\tan^{-1} x}}{(1+x^2)}dx=e^{\tan^{-1}x}\times\dfrac1{\sqrt{x^2+1}}+\int x(1+x^2)^{\tfrac{-3}2}e^{\tan^{-1}x}dx$$
$$=\dfrac{e^{\tan^{-1}x}}{\sqrt{x^2+1}}+\int\dfrac{{xe^{\tan^{-1}x}}}{(1+x^2)^{\frac32}}dx$$
But I don't know how to continue
 A: I wanted to figure out how the exponent of  $1+x^2$  is resolved to be $-\dfrac32$
$$\dfrac{d \left(\dfrac{e^{\tan^{-1}x}}{(1+x^2)^n}\right)}{dx}=\dfrac{e^{\tan^{-1}x}}{(1+x^2)^{n+1}}-2n\cdot\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^{n+1}}$$
Integrate both sides with respect to $x,$
$$\implies\dfrac{e^{\tan^{-1}x}}{(1+x^2)^n}+K=I_{n+1}-2nJ_{n+1}\  \ \ \  (1)$$  where $\displaystyle I_m=\int\dfrac{e^{\tan^{-1}x}}{(1+x^2)^m}\ dx\text{ and } J_m=\int\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^m}\ dx$
$$\dfrac{d \left(\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^n}\right)}{dx}=-(2n-1)\dfrac{e^{\tan^{-1}x}}{(1+x^2)^n}+\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^{n+1}}+2n\cdot\dfrac{e^{\tan^{-1}x}}{(1+x^2)^{n+1}}$$
Integrate both sides with respect to $x,$
$$\implies\dfrac{xe^{\tan^{-1}x}}{(1+x^2)^n}+C=-(2n-1)I_n+J_{n+1}+2nI_{n+1}\  \ \ \  (2)$$
If $2n-1=0,$ $(1),(2)$  becomes simultaneous equations for $$I_{3/2},J_{3/2}$$
Can you take it from here?
Had $2n-1$ been $\ne0,$ the problem would not have remained so elementary.
A: Let $I=\int\dfrac{e^{\tan^{-1} x}}{(1+x^2)^{\tfrac32}}dx$.  
Substitute $t=\tan ^{-1}x\implies dt=\frac 1{1+x^2}dx$. Now noting that $x=\tan t$, we have 
$I=\int e^t \frac 1{\sqrt{\sec^2t}}dt=\int e^t\cos t dt$ 
Can you take it from here?
