How to approach $\int\limits_{-\infty}^{a}\frac{\sin^{-1}e^x+\sec^{-1}e^{-x}}{(\tan^{-1}e^a+ \tan^{-1}e^x)(e^x+e^{-x})}\mathrm{d}x$ How to approach this integral?
$$\int\limits_{-\infty}^{a}\frac{\sin^{-1}e^x+\sec^{-1}e^{-x}}{(\tan^{-1}e^a+ \tan^{-1}e^x)(e^x+e^{-x})}\mathrm{d}x$$

 A: \begin{gather*}
Let\ I=\int ^{a}_{-\infty }\frac{\arcsin\left( e^{x}\right) +\ \arcsec\left( e^{-x}\right)}{\left( c+\arctan\left( e^{x}\right)\right)\left( e^{x} +e^{-x}\right)} dx\\
=\int ^{a}_{-\infty }\frac{\arcsin\left( e^{x}\right) +\arccos\left( e^{x}\right)}{\left( c+\arctan\left( e^{x}\right)\right)\left( e^{x} +e^{-x}\right)} dx\\
\left( As\ \arcsec( 1/t) =\arccos( t) \ for\ all\ t >0\ and\ e^{x}  >0\ for\ all\ x\right)\\
=\int ^{a}_{-\infty }\frac{\frac{\pi }{2}}{\left( c+\arctan\left( e^{x}\right)\right)\left( e^{x} +e^{-x}\right)} dx=\frac{\pi }{2}\int ^{e^{a}}_{0}\frac{e^{x} dx}{\left( c+\arctan\left( e^{x}\right)\right)\left( e^{2x} +1\right)}\\
Substitute\ e^{x} =t\\
e^{x} dx=dt\\
I=\frac{\pi }{2}\int ^{e^{a}}_{0}\frac{dt}{( c+\arctan( t))\left( t^{2} +1\right)}\\
Now\ let\ \arctan( t) =u\\
\frac{1}{t^{2} +1} dt=du\\
I=\frac{\pi }{2}\int ^{\arctan\left( e^{a}\right)}_{0}\frac{du}{( c+u)} =\frac{\pi }{2}\ln \mid u+c\mid =\frac{\pi }{2}\ln\left(\frac{\arctan\left( e^{a}\right)}{c} +1\right)\\
Remembering\ that\ c=\arctan\left( e^{a}\right) ,\\
I=\frac{\pi }{2}\ln( 2)
\end{gather*}
Hope this answers your question!
A: Note $ \sin^{-1}e^x+\cos^{-1}e^{x} =\frac\pi2$ and substitute  $y= \tan^{-1}e^a+ \tan^{-1}e^x$, $dy = \frac{dx}{e^x+e^{-x}}$ to integrate
$$\int_{-\infty}^{a}\frac{\sin^{-1}e^x+\sec^{-1}e^{-x}}{(\tan^{-1}e^a+ \tan^{-1}e^x)(e^x+e^{-x})}\mathrm{d}x
=\frac\pi2 \int_{ \tan^{-1}e^a}^{2\tan^{-1}e^a}\frac1ydy=\frac\pi2\ln2
$$
