How is this definite integral solved: $\int_{-\sqrt3}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx $?

Question

$$\int_{-\sqrt3}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx $$ Tried partially integrating, had no luck.. Any thoughts?

Answer 1

There is a trick.

You may just write $$\begin{align} \int_{-\sqrt3}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx &=\int_{-\sqrt3}^{0}{e^x\over(e^x+1)(x^2+1)}dx+\int_{0}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx\\\\ &=\int_0^{\sqrt3}{e^{-x}\over(e^{-x}+1)(x^2+1)}dx+\int_{0}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx\\\\ &=\int_0^{\sqrt3}{1\over(e^{x}+1)(x^2+1)}dx+\int_{0}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx\\\\ &=\int_0^{\sqrt3}{(1+e^{x})\over(e^{x}+1)(x^2+1)}dx\\\\ &=\int_0^{\sqrt3}{1\over x^2+1}dx\\\\ &=\left[\arctan x\right]_0^{\sqrt3}\\\\ &=\frac \pi3. \end{align}$$

Answer 2

There is another way to solve. Let $$ A=\int_{-\sqrt3}^{\sqrt3}{e^x\over(e^x+1)(x^2+1)}dx, B=\int_{-\sqrt3}^{\sqrt3}{1\over(e^x+1)(x^2+1)}dx. $$ Clearly $$ A+B=\int_{-\sqrt3}^{\sqrt3}\frac{1}{x^2+1}=2\frac{\pi}{3}.$$ Changing variable $x\to -x$, it is easy to get $A=B$. Thus $$ A=\frac{\pi}{3}. $$