Calculate $\int_{-1}^{1}\frac{t^{2}}{1+\exp(t)}dt$, whose indefinite integral is of non-elementary function 
Calculate $$\int_{-1}^{1}\frac{t^{2}}{1+\exp(t)}dt$$

I know an approach is using the change of variables $t=-u$, so we have the value for the integral $\frac{1}{3}$. How can I calculate that integral using other approach?
I thought about this  problem when observing this indefinite integral indefinite integral.
 A: Note
$$\frac1{1+e^t} = \frac12 -\frac12\tanh\frac t2$$
and the odd function $\tanh\frac t2$ does not survives the integration over $(-1,1)$. Thus, the integral reduces to
$\int_{-1}^{1}\frac{t^{2}}2 dt=\frac13$.
A: Let
$$I=\int_{-1}^{1}\frac{t^{2}}{1+e^t}dt=\int_{-1}^{0}\frac{t^{2}}{1+e^t}dt+\int_{0}^{1}\frac{t^{2}}{1+e^t}dt$$$$. $$
Under $t\to-t$,
$$\int_{-1}^{0}\frac{t^{2}}{1+e^t}dt=\int_0^1\frac{t^2}{1+e^{-t}}dt=\int_0^1\frac{t^2e^t}{1+et^t}dt$$
and hence
$$ I=\int_{-1}^{0}\frac{t^{2}}{1+e^t}dt+\int_{0}^{1}\frac{t^{2}}{1+e^t}dt=\int_0^1\frac{t^2+t^2e^t}{1+t^2}dt=\int_0^1t^2dt=\frac13. $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that
$\ds{\left.{1 \over \expo{t} + 1}\,
\right\vert_{\,t\ \not=\ 0} = \Theta\pars{-t} +
{\on{sgn}\pars{t} \over \expo{\verts{t}} + 1}}$ and equal to $\ds{1 \over 2}$ when $\ds{t = 0}$. $\ds{\Theta}$ is the Heaviside Theta or/and Step Function.
Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{-1}^{1}{t^{2} \over 1 + \expo{t}}\,\dd t} =
\int_{-1}^{0}t^{2}\,\dd t = \bbx{1 \over 3} \\ &
\end{align}
