What is the integral of (x^2+1)/(e^x+1) from -1 to 1?

[anemone]

Here is this week's POTW:

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Evaluate $$\int_{-1}^{1} \dfrac{x^2+1}{e^x+1}\,dx$$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to castor28 for his correct solution(Cool), which you can find below:

Spoiler

We write the expression as:
$$
\int_{-1}^0{\frac{x^2+1}{e^x+1}dx}+\int_0^1{\frac{x^2+1}{e^x+1}dx}
$$
We have:
\begin{align*}
\int_{-1}^0{\frac{x^2+1}{e^x+1}dx} &= -\int_1^0{\frac{x^2+1}{e^{-x}+1}dx}\\
&=+\int_0^1{\frac{x^2+1}{e^{-x}+1}dx}
\end{align*}
and the expression becomes:
$$
\int_0^1{(x^2+1)\left(\frac{1}{e^x+1}+\frac{1}{e^{-x}+1}\right)\,dx}
$$
On the other hand, we have:
$$
\frac{1}{e^x+1}+\frac{1}{e^{-x}+1} = \frac{e^x+e^{-x}+2}{1+e^x+e^{-x}+1}=1
$$
and we are left with:
$$
\int_0^1{(x^2+1)\,dx} = \left[\frac{x^3}{3}+x\right]_0^1 = {\bf\frac43}
$$