How to evaluate $\int \frac{\arctan x}{x\sqrt {1+x^2}}\exp\left(-\frac{\arctan x}{x}\right)\,dx$

How to evaluate $$\int \frac{\arctan x}{x\sqrt {1+x^2}}e^{-\arctan x/x} \, dx$$

I tried using substitution $x=\tan t,dx=\sec ^2tdt,$It becomes the following： $$\int\frac{t}{\sin t}e^{-t\cot t}dt$$

Then I got stuck，any hint of help is welcome.

• i think there is no solution in the known elementary functions – Dr. Sonnhard Graubner Sep 3 '17 at 11:51
• Thank you.Isn't there a elementary function? – JamesJ Sep 3 '17 at 12:03
• i think no try Wolfram Alpha – Dr. Sonnhard Graubner Sep 3 '17 at 12:03

We guess a primitive of the form $$\int\frac{t}{\sin t}e^{-t\cot t} dt = e^{-t\cot t}u$$ where $u$ is a function of $t$. Then \begin{aligned} &t\csc t{e^{ - t\cot t}} = ({e^{ - t\cot t}}u)' = (t{\csc ^2}t - \cot t){e^{ - t\cot t}}u + {e^{ - t\cot t}}u'\\ &\iff t\csc t = (t{\csc ^2}t - \cot t)u + u'\\ &\iff t\sin t = (t - \cos t\sin t)u + u'{\sin ^2}t \end{aligned}
One immediately notices that $u=\sin t$ is a solution.
Hence $$\int {\frac{t}{{\sin t}}{e^{ - t\cot t}}dt} = {e^{ - t\cot t}}\sin t + C$$