Compute $\lim_{n\to\infty}n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx$ Compute

\begin{equation}
\lim_{n\to\infty}n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx
\end{equation}

According to Wolfram Alpha, the limit is zero. I tried to make substitution $x=\frac{t}{n}$ and I got
\begin{equation}
\lim_{n\to\infty}\int_0^{\large\frac{1}{n}}\frac{t^n\ln^3t}{n^n+t^n}\ln\left(1-\frac{t}{n}\right)\,dx \to 0
\end{equation}
but I am not sure this approach is correct.
I also tried to make substitution $t=x^n$ and I got
\begin{equation}
\lim_{n\to\infty}n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx=\lim_{n\to\infty}\int_0^1\frac{t^{\large\frac{1}{n}}\ln^3t}{1+t}\ln\left(1-t^{\large\frac{1}{n}}\right)\,dt
\end{equation}
I used the bound $$\left|\int_0^1\frac{t^{\large\frac{1}{n}}\ln^3t}{1+t}\ln\left(1-t^{\large\frac{1}{n}}\right)\,dt\right|\leq\int_0^1\left|t^{\large\frac{1}{n}}\ln^3t\ln\left(1-t^{\large\frac{1}{n}}\right)\right|\,dt$$
because  $1+t\ge1$, so $\frac{1}{1+t}\leq1$. I can compute the integral using Taylor series for the logarithm but after taking the limit I got the result was infinity.
\begin{align}
\lim_{n\to\infty}n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx&=\lim_{n\to\infty}\int_0^1t^{\large\frac{1}{n}}\ln^3t\ln\left(1-t^{\large\frac{1}{n}}\right)\,dt\\
&=-\lim_{n\to\infty}\sum_{k=1}^\infty\frac{1}{k}\int_0^1t^{\large\frac{k+1}{n}}\ln^3t\,dt\\
&=\lim_{n\to\infty}\sum_{k=1}^\infty\frac{6}{k\left(\frac{k+1}{n}+1\right)^4}\to\infty
\end{align}
Could anyone here please help me? Any help would be greatly appreciated. Thank you.
 A: We will prove that the limit is $+\infty$.
Let $$\eqalign{I_n&=n^4\int_0^1\frac{x^n}{1+x^n}\ln^3 x\ln(1-x)dx\cr
J_n&=n^4\int_0^1x^n\ln^3 x\ln(1-x)dx}$$
Clearly $$\frac{1}{2}J_n\leq I_n\leq J_n$$
because $\frac{1}{2}\leq\frac{1}{1+x^n}\leq1$ and $\ln^3x\ln(1-x)\geq0$ for $0<x<1$.
So, let us consider $J_n$.
We have
$$-\ln(1-x)=\sum_{k=1}^\infty\frac{x^k}{k}$$
and
$$\int_0^1x^{m}(-\ln x)^3dx=\frac{6}{(m+1)^4}$$
Combining these two properties we see that
$$
J_n=\sum_{k=1}^\infty\frac{6n^4}{k(k+1+n)^4}
$$
for a given $m\geq1$ we have
$$
J_n\geq\sum_{k=1}^m\frac{6n^4}{k(k+1+n)^4}
$$
hence
$$\liminf_{n\to\infty}J_n\geq 6 \sum_{k=1}^m\frac{1}{k}$$
but $m$ is arbitrary and $\sum \frac{1}{k}$ is divergent, So
$$\liminf_{n\to\infty}J_n=+\infty$$
that is $\lim\limits_{n\to\infty}J_n=+\infty$, and consequently $\lim\limits_{n\to\infty}I_n=+\infty$.
A: For the time being, the only thing I can tell you is that
$$\begin{equation}
I_n=n^4\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\,dx
\end{equation}$$
seems to be an increasing function of $n$ (this has been done numerically): $$I_{10}=4.92397$$ $$I_{100}=18.2376$$ $$I_{1000}=31.7992$$ $$I_{10000}=44.97982$$ $$I_{100000}=58.0781$$
A: Omran Kouba has correctly answered the OP question. 
Let's just see how the integral behaves as $n$ is great.
As $n \rightarrow +\infty$, we have

$$
\lim_{n \to +\infty} \,\color{blue}{\frac{n^4}{\ln n}}\!\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\:\mathrm dx = \color{blue}{\frac{7\pi^4}{120}}. \tag1
$$

Moreover,

$$
\int_0^1\frac{x^n\ln^3x}{1+x^n}\ln(1-x)\:\mathrm dx = \frac{7\pi^4}{120}\frac{\ln n}{n^4}-\frac{45\:\zeta(5)}{2}\frac{\ln n}{n^5}+\mathcal{O}\left(\frac{\ln n}{n^6}\right). \tag2
$$

Recall that
$$
\int_0^1x^s\ln(1-x) \:\mathrm d x =- \frac{1}{s+1}\left(\gamma+\psi(s+2)\right), \, s>-1, \tag3
$$ where $\psi:=\Gamma'/\Gamma$ is the digamma function, as may be easily seen by expanding the logarithmic term and integrating termwise.
Then, using $\displaystyle \ln^3x=\partial_s^3 \left. \left(x^{s}\right) \right|_{s=0}$, one may write 
$$
\begin{align}
\int_0^1\frac{x^{n}\ln^3x}{1+x^n}\ln(1-x)\:\mathrm dx & = \partial_s^3 \left(\left. \int_0^1\frac{x^{n+s}}{1+x^n}\ln(1-x)\:\mathrm dx\right)\right|_{s=0}
 \\\\
& =\partial_s^3 \left(\left.\sum_{k=0}^{\infty}(-1)^k\int_0^1x^{n+s}x^{kn}\ln(1-x)\:\mathrm dx\right)\right|_{s=0}
\\\\
& =\partial_s^3 \left(\left.\sum_{k=1}^{\infty}\frac{(-1)^k}{nk+s+1}\left(\gamma+\psi(nk+s+2)\right)\right)\right|_{s=0}
\\\\
& =\partial_s^3 \left.\left(-\frac{\ln n}{n^4}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^4}+\frac{4\ln n}{n^5}\sum_{k=1}^{\infty}\frac{(-1)^k}{k^5}+\mathcal{O}\left(\frac{\ln n}{n^6}\right) \right)s^3\right|_{s=0}
\end{align}
$$ where we have used the uniform convergence on $s \in [0,1]$, and the asymptotic expansion of the digamma function, as $M \rightarrow +\infty$,
$$
\psi(M) = \ln M-\frac{1}{2M}-\frac{1}{12M^2}+\frac{1}{120M^4}-\frac{1}{252M^6}+\mathcal{O}\left(\frac{1}{M^8}\right) ,
$$
to obtain $(2)$, deducing $(1)$.
A: The comment box does not seem to like this equation! So if this is completely taking you in the wrong direction then please comment below and I will remove.
Would this be of benefit here
$$
\lim_{n\rightarrow \infty}\left[\lim_{\alpha\rightarrow 0,\beta\rightarrow 0}\dfrac{\partial^3}{\partial \alpha^3}\dfrac{\partial}{\partial \beta}n^4\int_{0}^{1}\dfrac{x^{\alpha}\left(1-x\right)^{\beta}x^n}{1+x^n}dx\right]
$$
