Im very weak in this concept, so please, if the following is not the cleanest method to solve the problem, then comment below about it.

$I(k) = \int_0^{\infty} \frac{x^2}{1+e^{kx}}dx$

$I'(k) = \int_0^{\infty} \frac{-x^3e^{kx}}{(1+e^{kx})^2}dx$

Let $\displaystyle 1+e^{kx} = t$

$I'(k) = \frac{-1}{k^4} \int_2^{\infty} \frac{(\ln(t-1))^3}{t^2}dt$

This integral can be easily evaluated using the expansion of $\displaystyle\ln(t-1)$ around the point $\displaystyle t = 2$

$I'(k) = \frac{-1}{k^4} \frac{9}{2} \zeta(3)$

$I(k) = \frac{9}{2}\zeta(3) \int\frac{-1}{k^4}$

$I(k) = \frac{9}{2}\zeta(3) \frac{1}{3k^3}$

Substituting $\displaystyle k=1$ to get the required integral, we have,

$I(1) = \frac{3}{2}\zeta(3) = \boxed{1.80308}$

$$ This is another approach in addition to Anish's solution. From the formulation of Dirichlet eta function we have: $$ $\eta(s)=\left(1-2^{1-s}\right)\zeta(s)$ and $\eta(s)=\frac{1}{\Gamma(s)}\int_0^\infty\frac{x^{s-1}}{1+e^x}\,dx.$ $$ where $\Gamma(s)$ is gamma function and $\Gamma(s)=(s-1)!$ for $s\in\mathbb{Z}_+$ .

Therefore, the solution of the integral becomes straightforward by using Dirichlet eta function. $$ $\begin{aligned} \int_0^\infty\frac{x^2}{1+e^x}\,dx &=(1-2^{-2})\Gamma(3)\zeta(3)\\ &=\frac{3}{4}\cdot 2!\cdot1.20205\\ &=\boxed{\color{#3D99F6}{1.80308}} \end{aligned}$ $$

---

$\Large\color{#3D99F6}{\text{\# }\mathbb{Q.E.D.}\text{ \#}}$

First of all, note that

$\frac{1}{1+e^x} = \frac{1}{e^x} \left ( \frac{1}{1+\frac{1}{e^x}} \right ) = \frac{1}{e^x} \sum_{k=0}^{\infty} e^{-kx} (-1)^k = \sum_{k=0}^{\infty} (-1)^k e^{-x(k+1)} \;\;\; |e^x|<1$

Then

$\int_{0}^{\infty} \frac{x^2}{1+e^x} \; dx = \sum_{k=0}^{\infty} (-1)^k \int_{0}^{\infty} x^2e^{-x(k+1)} \; dx$

Now, integration by parts will give us

$\int_{0}^{\infty} x^2e^{-x(k+1)} \; dx = - \left (\frac{x^2e^{-x(k+1)}}{k+1} \right )_{0}^{\infty} + 2 \int_{0}^{\infty}x \; \frac{e^{-x(k+1)}}{k+1} \; dx$

And

$\int_{0}^{\infty}x \; \frac{e^{-x(k+1)}}{k+1} \; dx = - \left ( \frac{xe^{-x(k+1)}}{k+1} \right )_{0}^{\infty} + \int_{0}^{\infty} \frac{e^{-x(k+1)}}{(k+1)^2} \; dx$

Therefore

$\int_{0}^{\infty} x^2e^{-x(k+1)} \; dx = - \left (\frac{x^2e^{-x(k+1)}}{k+1} \right )_{0}^{\infty} - 2 \left ( \frac{xe^{-x(k+1)}}{k+1} \right )_{0}^{\infty} + 2 \int_{0}^{\infty} \frac{e^{-x(k+1)}}{(k+1)^2} \; dx$

Now using L'Hospital

$\lim_{x\rightarrow \infty} \frac{x^2}{e^{x(k+1)}} = \lim_{x\rightarrow \infty} \frac{2x}{(k+1)e^{x(k+1)}} = \frac{2}{(k+1)^2} \lim_{x\rightarrow \infty} \frac{1}{e^{x(k+1)}} = 0$

And therefore

$\int_{0}^{\infty} x^2e^{-x(k+1)} \; dx = 2 \int_{0}^{\infty} \frac{e^{-x(k+1)}}{(k+1)^2} \; dx = -2\left ( \frac{e^{-x(k+1)}}{(k+1)^3} \right )_{0}^{\infty} = 2\left ( \frac{1}{k+1} \right )^3$

Thus the original integral is

$\int_{0}^{\infty} \frac{x^2}{1+e^x} \; dx = 2\sum_{k=0}^{\infty} (-1)^k \left ( \frac{1}{k+1} \right )^3 = 2 \sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n^3} = \left ( \frac{7}{8} - \frac{1}{8} \right ) \zeta(3) \approx 1.803$