Proving: $\int_0^1 \int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$ 
Proving:$$\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\frac{225}{2}\zeta(5)$$

I tried using variable switching
$\ln(xy)=t$  But I did not reach any results after the calculation
\begin{align*}
k&=\displaystyle\int_0^1\displaystyle\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dx\\
&=\displaystyle\int_0^1\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4e^t}{(1+e^t)^2y^2}dtdy\\
&=\displaystyle\int_0^1\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4e^t}{y^2(1+e^t)}\displaystyle\sum_{n=0}^{\infty}(-e^t)dtdy\\
&=\displaystyle\int_0^1\frac{1}{y^2}\left(\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{-\infty}^{\ln(y)}\frac{t^4(-e^{2t})}{1+e^t}dt\right)dy\\
&=\displaystyle\int_0^1\frac{1}{y^2}\left(\displaystyle\sum_{n=0}^{\infty}\displaystyle\int_{\ln(y)}^{\infty}\frac{t^4e^{2t}}{1+e^t}dt\right)dy\\
\end{align*}
 A: Starting off after your first substitution, notice that $$\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}, \text{ for } x \in(-1,1)$$
Since the domain of $x,y$ is $(0,1)$, we can write $$\frac{e^t}{(1+e^t)^2}=-\sum_{n=1}^{\infty} {(-1)}^n n e^{tn}$$
In addition, you made a slight error when calculating $dt$ I suppose.  It should be $y$ not $y^2$ in the denominator.
$$\int_0^1 \int_0^{\ln{y}} \frac{t^4}{y}\left( -\sum_{n=1}^{\infty} {(-1)}^n n e^{tn}\right) \; dt \; dy$$
Because the summation converges, we can interchange the summation and integral sign from Fubini's theorem:
\begin{align}
k &= -\sum_{n=1}^{\infty} {(-1)}^n n \int_0^1 \frac{1}{y} \int_0^{\ln{y}} t^4 e^{tn}\; dt \; dy \\
&\overset{\text{IBP}}= -\sum_{n=1}^{\infty} {(-1)}^n n \int_0^1 \frac{y^{n-1} \left(n^4  \ln^4{y}-4n^3\ln^3{y}+12n^2\ln^2{y}-24n\ln{y}+25\right)}{n^5} \; dy \\
&\overset{\text{IBP}}= -\sum_{n=1}^{\infty} {(-1)}^n n \cdot \frac{120}{n^6} \\
&= 120\sum_{n=1}^{\infty} \frac{{(-1)}^{n+1}}{n^5} \\
&= \boxed{\frac{225}{2}\zeta(5)}  \\
\end{align}
A: \begin{align}J&=\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy\\
&=\int_0^1 \frac{1}{x}\left(\int_0^x \frac{\ln^4 u}{(1+u)^2 du}\right)\\
&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^4 u}{(1+u)^2} du\right)\ln x\right]_0^1-\int_0^1 \frac{\ln^5 x}{(1+x)^2}dx\\
&=-\int_0^1 \frac{\ln^5 x}{(1+x)^2}dx\\
&\overset{\text{IBP}}=\left[\ln^5 x\left(\frac{1}{1+x}-1\right)\right]_0^1-5\int_0^1 \frac{\left(\frac{1}{1+x}-1\right)\ln^4 x}{x}dx\\
&=5\int_0^1 \frac{\ln^4 x}{1+x}dx\\
&=5\left(\int_0^1 \frac{\ln^4 x}{1-x}dx-\int_0^1 \frac{2t\ln^4 t}{1-t^2}dt\right)\\
&\overset{x=t^2}=5\left(\int_0^1 \frac{\ln^4 x}{1-x}dx-\frac{1}{16}\int_0^1 \frac{\ln^4 x}{1-x}dx\right)\\
&=\frac{5\times 15}{16}\int_0^1 \frac{\ln^4 x}{1-x}dx\\
&=\frac{5\times 15}{16}\times 24\zeta(5)\\
&=\boxed{\frac{225}{2}\zeta(5)}\\
\end{align}
I assume that,  $\displaystyle \int_0^1 \frac{\ln^4 x}{1-x}dx=24\zeta(5)$
A: Start with $xy=t$ we have
$$\int_0^1\int_0^1\frac{\ln^4(xy)}{(1+xy)^2}dxdy=\int_0^1\int_0^x\frac{\ln^4t}{x(1+t)^2}dxdt$$
$$=\int_0^1\frac{\ln^4t}{(1+t)^2}\left(\int_t^1\frac{dx}{x}\right)dt=-\int_0^1\frac{\ln^5t}{(1+t)^2}dt$$
$$=\sum_{n=1}^\infty (-1)^n n\int_0^1 t^{n-1}\ln^5tdt=120\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^5}$$
$$=120\eta(5)=120\cdot\frac{15}{16}\zeta(5)=\frac{225}{2}\zeta(5)$$
Note that $\eta(s)=(1-2^{1-s})\zeta(s)$ is the Eta function.
