# Integral $\int_{-1}^{1} \frac{1}{(e^x+1)(x^2+1)}$ [duplicate]

I do not know how to approach this integration problem. $$\int_{-1}^{1} \frac{1}{(e^x+1)(x^2+1)}$$ I tried some trigonometric change of variable and also tried to break it into two fractions but I failed!

• How about breaking the integral into two integrals, over the non-negative $x$ and the non-positive $x$, and combining them suitably? – Mindlack Dec 13 '18 at 19:36
• Another one math.stackexchange.com/q/1805550/42969 – both found with Approach0 – Martin R Dec 13 '18 at 19:48
• $\displaystyle\mathrm{f}\left(\,{x}\,\right) \equiv {1 \over \mathrm{e}^{x} + 1} = \Theta\left(\,{-x}\,\right) + \,\mathrm{sgn}\left(\,{x}\,\right)\,\mathrm{f}\left(\,{\left\vert\, x\,\right\vert}\,\right).\quad\Theta$ is the Heaviside Step Function. – Felix Marin Dec 13 '18 at 21:31

$$\displaystyle I=\int_{-1}^{1}\frac{dx}{(e^x+1)(x^2+1)}$$
We know that $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
$$\displaystyle\implies I=\int_{-1}^1\frac{dx}{(e^{-x}+1)(x^2+1)}=\int_{-1}^1\frac{e^xdx}{(e^x+1)(x^2+1)}$$
$$\displaystyle\implies I+I=2I=\int_{-1}^1\frac{(e^x+1)dx}{(e^x+1)(x^2+1)}=\int_{-1}^1\frac{dx}{x^2+1}=2\int_0^1\frac{dx}{x^2+1}=2\tan^{-1}(1)$$
which gives $$I=\pi/4$$