Can this integral be evaluated properly by using highschool integration skills ?
$$\int_{-1}^{1}\dfrac{x^2}{e^x+1}\,dx$$
(Original image at https://i.stack.imgur.com/AqOsX.png)
Judging from what I get from WolframAlpha, the primitives involve dilogarithms and other than that they look pretty messy as well. Can you suggest some kind of a solution, or maybe an interesting approach ?
Thanks !
Let $I = \displaystyle\int_{-1}^{1}\dfrac{x^2}{e^x+1}\,dx$. Use the substitution $x = -u$, $dx = -du$ to get:
$I = -\displaystyle\int_{1}^{-1}\dfrac{(-u)^2}{e^{-u}+1}\,du = \int_{-1}^{1}\dfrac{u^2}{e^{-u}+1}\,du = \int_{-1}^{1}\dfrac{u^2e^u}{1+e^u}\,du$.
Therefore, $2I = I+I = \displaystyle\int_{-1}^{1}\dfrac{x^2}{e^x+1}\,dx + \int_{-1}^{1}\dfrac{u^2e^u}{1+e^u}\,du$
$= \displaystyle\int_{-1}^{1}\left[\dfrac{x^2}{e^x+1} + \dfrac{x^2e^x}{1+e^x}\right]\,dx = \int_{-1}^{1}\dfrac{x^2(1+e^x)}{1+e^x}\,dx = \int_{-1}^{1}x^2\,dx = \dfrac{2}{3}$.
Since $2I = \dfrac{2}{3}$, we have $I = \dfrac{1}{3}$.
