Ndtyjky

First off, notice the integrand is even, so we have $$ int_{-infty}^infty frac{x^2 e^x}{(e^x+1)^2}dx = 2int_{0}^infty frac{x^2 e^x}{(e^x+1)^2}dx.$$ Then we can expand $$ frac{1}{(1+x)^2} = sum_{n=0}^infty (-1)^{n}(n+1) x^n $$ and write $$ 2int_{0}^infty frac{x^2 e^x}{(e^x+1)^2}dx=\ =2int_{0}^infty frac{x^2 e^{-x}}{(e^{-x}+1)^2}dx \ = 2 int_0^infty x^2e^{-x}sum_{n=0}^infty (-1)^n (n+1) e^{-nx}\=2sum_{n=0}^infty (-1)^n(n+1) int_0^infty x^2 e^{-(n+1)x}dx\=4sum_{n=0}^infty frac{(-1)^n}{(n+1)^2}.$$ Then we have $$ sum_{n=0}^infty frac{(-1)^n}{(n+1)^2} = 1-frac{1}{2^2} + frac{1}{3^2}ldots = (1-frac{2}{2^2})(1+frac{1}{2^2} + frac{1}{3^2}ldots) = frac{pi^2}{12}$$

Edit
Realized this can be simplified somewhat by first doing an integration by parts $$ 2int_0^infty frac{x^2 e^x}{(e^x+1)^2}dx = 4int_0^infty frac{x}{e^x+1}dx$$ followed by a similar series expansion. Additionally, this solution somewhat 'misses the point' relative to contour integration approaches since that's one of the slicker ways to get $sum_{n}1/n^2=pi^2/6$ in the first place (and also transformation from the sums to integrals like this are the source of many zeta function identities).

Substitute $e^x=t$ we get $$I=int_0^{infty} frac {ln^2t dt}{(1+t)^2}$$

Let $$J(a)=int_0^{infty} frac {t^{a-1}dt}{(1+t)^2}$$

So we need to find $J''(1)$

Notice that $$J(a)=B(a,2-a)=frac {Gamma(a)Gamma(2-a)}{Gamma(2)}$$

Where $B(x,y)$ is Standard beta function and $Gamma(x)$ is the Gamma function

Using the properties of Gamma function and Euler's reflection formula we have $$J(a)=Gamma(a) cdot(1-a)Gamma(1-a)=pi frac {1-a}{sin pi a}$$

Differentiating twice w.r.t $a$ and taking limit $ato 1$ gives the desired result

You have
$$
int_{-infty}^inftyfrac{x^2 e^x}{(e^x+1)^2},mathrm{d}x=
int_{-infty}^inftyfrac{x^2}{2(cosh x+1)},mathrm{d}x
$$
So integrate $dfrac{z^3}{1+cosh z}$ around rectangle $pm Rpm2pi i$, we have encircled a simple pole at $pi i$ with residue $6pi^2$. The vertical sides contribution is $to 0$ as $Rtoinfty$ because of cosh in the denominator, and the horizontal contribution
$$
int_{-infty}^inftyfrac{x^3}{1+cosh x},mathrm{d}x+int_infty^{-infty}frac{(x+2pi i)^3}{1+underbrace{cosh(x+2pi i)}_{=cosh x}},mathrm{d}x
$$
whose imaginary part is
$$
2piint_infty^{-infty}frac{3x^2-4pi^2}{1+cosh(x)},mathrm{d}x.
$$
So
$$
3int_infty^{-infty}frac{x^2}{1+cosh(x)},mathrm{d}x
-4pi^2int_infty^{-infty}frac{1}{1+cosh(x)},mathrm{d}x
=6pi^2
$$
i.e.,
begin{align}
int_{-infty}^inftyfrac{x^2}{2(1+cosh(x))},mathrm{d}x
&=-pi^2
+frac43pi^2int_{-infty}^inftyfrac{1}{2(1+cosh(x))},mathrm{d}x
\
&=-pi^2+frac43pi^2=frac13pi^2
end{align}
since
$$
int_{-infty}^inftyfrac{1}{2(cosh x+1)},mathrm{d}x=int_{-infty}^inftyfrac14operatorname{sech}^2frac x2,mathrm{d}x=left[frac12tanhfrac x2right]_{-infty}^infty=1.
$$

Here we use Feynman's Trick to facilitate analysis. Proceeding, we have

$$begin{align}
I&=int_{-infty}^infty frac{x^2e^x}{(e^x+1)^2},dx\\
&=2int_0^infty frac{x^2e^x}{(e^x+1)^2},dx\\
&=-2left.left(frac{d}{dy}int_0^infty frac{x^2}{ye^x+1},dxright)right|_{y=1}\\
&=-2left.left(frac{d}{dy}int_0^infty frac{x^2e^{-x}}{e^{-x}+y},dxright)right|_{y=1}\\
&=-2left.frac{d}{dy}left(frac1yint_0^infty frac{x^2e^{-x}}{e^{-x}/y+1},dxright)right|_{y=1}\\
&=-2left.frac{d}{dy}left(sum_{n=0}^infty frac{(-1)^n}{y^{n+1}}int_0^infty x^2e^{-(n+1)x},dxright)right|_{y=1}\\
&=-4left.frac{d}{dy}left(sum_{n=0}^infty frac{(-1)^n}{y^{n+1}(n+1)^3}right)right|_{y=1}\\
&=4sum_{n=1}^infty frac{(-1)^{n+1}}{n^2}\\
&=4frac{pi^2}{12}\\
&=frac{pi^2}3
end{align}$$

my first thought was trying substitution:
$$I=int_{-infty}^inftyfrac{x^2e^x}{(e^x+1)^2}dx$$
$u=e^x+1,,dx=frac{du}{e^x}$
$$I=int_1^inftyfrac{x^2}{u^2}du=int_1^inftyfrac{ln^2(u-1)}{u^2}du$$
if we now let:
$$I(a)=int_1^inftyfrac{ln(au-1)}{u^2}du$$
then:
$$I''(a)=int_1^inftyln(au-1)du$$
$v=au-1,,du=frac{dv}{a}$
$$I''(a)=frac1aint_a^inftyln(v)dv$$
although this leaves us with a divergent integral, which does not help.

Another approach:
$$I=int_{-infty}^inftyfrac{x^2e^x}{(e^x+1)^2}dx=int_{-infty}^inftyfrac{x^2[(e^x+1)-1]}{(e^x+1)^2}dx=int_{-infty}^inftyfrac{x^2}{e^x+1}dx-int_{-infty}^inftyfrac{x^2}{(e^x+1)^2}dx$$

The integral you are interested in can be rewritten in the form: $$int_{-infty}^{infty}frac{x^{2}}{2 cosh(x)+2} dx$$

Several techniques for solving integrals similar to this can be found at:

Improper Integral of $x^2/cosh(x)$

One of the standard solutions for evaluating difficult integrals is to transform the integrand into an infinite sequence and then integrating termwise. Indeed, if we start off with the geometric sum$$sumlimits_{ngeq0}x^n=frac 1{1-x}$$multiply both sides by $x$ and differentiating with respect to $x$, it's easy to see that$$sumlimits_{ngeq0}(n+1)x^n=frac 1{(1-x)^2}$$Replace $x$ with $-e^{-x}$ in the problem and calling the integral $mathfrak{I}$, the problem transforms into$$begin{align*}mathfrak{I} & =2sumlimits_{ngeq0}(-1)^n(n+1)intlimits_{0}^{infty}mathrm dx,x^2 e^{-(n+1)x}end{align*}$$The remaining integral can be evaluated using integration by parts on $u=x^2$ to get$$mathfrak I=4sumlimits_{ngeq0}frac {(-1)^{n}}{(n+1)^2}=4sumlimits_{ngeq1}frac {(-1)^{n-1}}{n^2}$$The inner sum is equal to $pi^2/12$ and is due to the identity$$eta(s)=(1-2^{1-s})zeta(s)$$Where$$eta(s)=sumlimits_{ngeq1}frac {(-1)^{n-1}}{n^s}$$$$zeta(s)=sumlimits_{ngeq1}frac 1{n^s}$$Since $zeta(2)=pi^2/6$, the result follows immediately$$intlimits_{-infty}^{infty}mathrm dx,frac {x^2e^{x}}{(1+e^x)^2}color{blue}{=frac {pi^2}{3}}$$

We could even compute the antiderivative since
$$I=intfrac{x^2 e^x}{(e^x+1)^2},dx=x left(frac{e^x x}{e^x+1}-2 log left(e^x+1right)right)-2 text{Li}_2left(-e^xright)$$ making
$$J=int_0^pfrac{x^2 e^x}{(e^x+1)^2},dx=frac{e^p p^2}{e^p+1}-2 text{Li}_2left(-e^pright)-2 p log
left(e^p+1right)-frac{pi ^2}{6}$$ Expanding as series for large values of $p$
$$J=left(2-frac{1}{e^p+1}right) p^2+frac{1}{2} e^{-2 p} left(1-4
e^pright)+frac{pi ^2}{6}-2 p log left(e^p+1right)+cdots$$ which shows the limit and which is a quite good approximation even for small values of $p$
$$left(
begin{array}{ccc}
p & text{approximation} & text{exact} \
1 & 0.08137802971721 & 0.07217327763488 \
2 & 0.39889758755901 & 0.39838536515902 \
3 & 0.82824233816028 & 0.82821565813003 \
4 & 1.17549173764877 & 1.17549038617232 \
5 & 1.39700611481663 & 1.39700604709488 \
6 & 1.52125697930945 & 1.52125697592972 \
7 & 1.58570867980301 & 1.58570867963459 \
8 & 1.61743428754382 & 1.61743428753543 \
9 & 1.63247105478514 & 1.63247105478472 \
10 & 1.63939550316470 & 1.63939550316468
end{array}
right)$$