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

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The integral of (x^2+1)/(e^x+1) from -1 to 1 is evaluated in this week's Problem of the Week. Participants are encouraged to follow the guidelines for submissions and solutions. Castor28 provided the correct solution, which is highlighted in the discussion. The evaluation of this integral involves techniques from calculus, particularly integration methods. The thread emphasizes the importance of engaging with the community for problem-solving.
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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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Congratulations to castor28 for his correct solution(Cool), which you can find below:
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:
$$
\int_0^1{(x^2+1)\,dx} = \left[\frac{x^3}{3}+x\right]_0^1 = {\bf\frac43}
$$
 
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