Solving definite integral from -1 to 1 Please help me with this. I can't prove the result. Tried integral by parts or notations, nothing working
$$\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx$$

Let $t=-x$, then
\begin{align*}
\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
&=\int_{1}^{-1}{\frac{t^2}{e^{-t}+1}}(-dt)\\
&=\int_{-1}^{1}{\frac{t^2}{e^{-t}+1}}dt\\
&=\int_{-1}^{1}{\frac{x^2e^x}{e^x+1}}dx.
\end{align*}
\begin{align*}
\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
&=\frac{1}{2}\left(\int_{-1}^{1}{\frac{x^2}{e^x+1}}dx
+\int_{-1}^{1}{\frac{x^2e^x}{e^x+1}}dx\right)\\
&=\frac{1}{2}\int_{-1}^{1}x^2dx=\int_{0}^{1}x^2dx\\
&=\frac{1}{3}.
\end{align*}
