Integrating $ \int_0^\infty \frac{x^5}{e^x+1} \, dx $

$$I=\int _{ 0 }^{ \infty }{ \cfrac { { x }^{ 5 } }{ { e }^{ x }+1 } dx= } \int _{ 0 }^{ \infty }{ \cfrac { { e }^{ -x } }{ { 1+e }^{ -x } } { x }^{ 5 }dx } =\int _{ 0 }^{ \infty }{ \left( \sum _{ r=1 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx } } { x }^{ 5 } \right) dx } \\ =\sum _{ r=1 }^{ \infty }{ \left\{ \int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } \right\} }$$

Let $rx=t$: $$\int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } ={ \left( -1 \right) }^{ r-1 }\int _{ 0 }^{ \infty }{ { e }^{ -t }{ \left( \cfrac { t }{ r } \right) }^{ 6-1 }\cfrac { dt }{ r } } =\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \Gamma (6)=120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \\ \\ \therefore I=\sum _{ r=1 }^{ \infty }{ 120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } } =120\eta \left( 6 \right) =120\left\{ \left( 1-{ 2 }^{ 1-6 } \right) \right\} \zeta \left( 6 \right) \\ \\ =120\times \cfrac { 31 }{ 32 } \zeta \left( 6 \right) =\cfrac { 465 }{ 4 } \zeta \left( 6 \right)$$

$$\zeta \left( 6 \right) =\zeta \left( 2\times 3 \right) =\cfrac { { \left( -1 \right) }^{ 3+1 }{ B }_{ 6 }{ \left( 2\pi \right) }^{ 6 } }{ 2\left( 6 \right) ! } =\cfrac { { \pi }^{ 6 } }{ 945 }$$ Note that $B_6$ refers to the $6$th Bernoulli Number $\left(B_6=\frac{1}{42}\right)$:

$$\therefore I=\cfrac { 465 }{ 4 } \times \cfrac { { \pi }^{ 6 } }{ 945 } =\cfrac { { 31\pi }^{ 6 } }{ 252 }$$

Here is yet another slightly different approach. Here we enforce the substitution $e^{-x}\to x$ and obtain

$$\begin{align} \int_0^{\infty} \frac{x^5}{1+e^{x}}dx&=-\int_0^1 \frac{\left(\log x\right)^5}{1+x}dx\\\\ &=-\frac{d^5}{dy^5}\left.\left(\int_0^1\frac{x^y}{1+x}dx\right)\right|_{y=0} \end{align}$$

Expanding $\frac{1}{1+x}$ and integrating term by term reveals

$$\begin{align} -\frac{d^5}{dy^5}\left.\left(\int_0^1\frac{x^y}{1+x}dx\right)\right|_{y=0}&=-\frac{d^5}{dy^5}\left.\left(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n+y}\right)\right|_{y=0}\\\\ &=(5!)\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^6}\\\\ &=(5!)\eta(6)\\\\ &=(120)\frac{31\pi^6}{30240}\\\\ &=\frac{31\pi^6}{252} \end{align}$$

which agrees with the results reported by others as expected!

The key is to express the integrand as the sum of a geometric series, and then switch the order of the integral and the summation.

$\displaystyle\int_{0}^{\infty}\dfrac{x^5}{e^x+1}\,dx$ $= \displaystyle\int_{0}^{\infty}\dfrac{x^5e^{-x}}{1+e^{-x}}\,dx$ $= \displaystyle\int_{0}^{\infty}\sum_{n = 1}^{\infty}x^5e^{-x}(-e^{-x})^{n-1}\,dx$ $= \displaystyle\int_{0}^{\infty}\sum_{n = 1}^{\infty}(-1)^{n-1}x^5e^{-nx}\,dx$ $= \displaystyle\sum_{n = 1}^{\infty}(-1)^{n-1}\int_{0}^{\infty}x^5e^{-nx}\,dx$ $= \displaystyle\sum_{n = 1}^{\infty}(-1)^{n-1}\int_{0}^{\infty}\left(\dfrac{u}{n}\right)^5e^{-u}\cdot \dfrac{1}{n}\,du$ $= \displaystyle\sum_{n = 1}^{\infty}\dfrac{(-1)^{n-1}}{n^6}\int_{0}^{\infty}u^5e^{-u}\,du$ $= \displaystyle\left(\sum_{n = 1}^{\infty}\dfrac{(-1)^{n-1}}{n^6}\right)\cdot \left(\int_{0}^{\infty}u^5e^{-u}\,du\right)$ $= \left(1-\dfrac{2}{2^6}\right)\zeta(6) \cdot \Gamma(6)$ $= \dfrac{31}{32} \cdot \dfrac{\pi^6}{945} \cdot 120 = \dfrac{31\pi^6}{252}$.

Here, we have used the following formulas involving the Gamma function, and the Riemann Zeta function: $\Gamma(n+1) = \displaystyle\int_{0}^{\infty}x^ne^{-x}\,dx$ and $(1-2^{1-s})\zeta(s) = \displaystyle\sum_{n = 1}^{\infty}\dfrac{(-1)^{n-1}}{n^s}$.