The value of following integral $$\frac {x^2\cos (x)}{1+e^x} dx$$ from $[-\pi,\pi] $ ? Now I converted to $$R\frac{e^{ln (x^2)+ix}}{1+e^x}dx $$ wherw R is real part now it isnt an even or odd so those manipulations are useless. $1+e^x=u $ thus $e^xdx=du $ but from here I can go nowhere.

Following @Sangchul Lee's  comment (although the idea is pretty straight forward to someone who is aware of basic calculus technics) we have:
\begin{align*}
\int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^x+1} \, {\rm d}x &\overset{u=-x}{=\! =\! =\! =\!} \int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^{-x} +1} \, {\rm d}x \\ 
 &= \int_{-\pi}^{\pi} \frac{x^2 \cos x e^x}{1+e^x} \, {\rm d}x\\ 
 &= \frac{1}{2}\left ( \int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^x+1} \, {\rm d}x  + \int_{-\pi}^{\pi} \frac{x^2 \cos x e^x}{e^x+1} \right ) \, {\rm d}x\\ 
 &= \frac{1}{2} \int_{-\pi}^{\pi} \frac{x^2 \cos x \left ( e^x+1 \right )}{e^x+1} \, {\rm d}x \\
 &= \frac{1}{2} \int_{-\pi}^{\pi} x^2 \cos x \, {\rm d}x \\
 &= -2 \pi 
\end{align*}
The last integral is dealt with integration by parts twice and easy quite easy.
