Is there any common method to evaluate $\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t} \, dt \right)$? So is there any common method to solve this kind of problem?

Calculate that $$\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t} \, dt \right)$$

Before this I have seen a similar problem which was solved by using digamma function or beta function. So is that mean we can always use special functions to solve them?
 A: Note that
$$\frac{\ln^4 x}{4}=\int_{1}^{x}\frac{\ln^3 t}{t}dt.$$
So
\begin{align*}
\frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t}dt
  &=\int_{1}^{x}\frac{\ln^3 t}{t}dt-\int_0^x \frac{\ln^3 t}{1+t}dt\\
  &=\int_{1}^{x}\frac{\ln^3 t}{t}dt-\int_0^1 \frac{\ln^3 t}{1+t}dt-\int_1^x \frac{\ln^3 t} 
     {1+t}dt\\
  &=\int_{1}^{x}\frac{\ln^3 t}{t(1+t)}dt-\int_0^1 \frac{\ln^3 t}{1+t}dt.
\end{align*}
Then
$$\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t}dt \right)
=\int_{1}^{+\infty}\frac{\ln^3 t}{t(1+t)}dt-\int_0^1 \frac{\ln^3 t}{1+t}dt.$$
By variable substitution $t=\dfrac{1}{s}$,
$$\int_{1}^{+\infty}\frac{\ln^3 t}{t(1+t)}dt=-\int_0^1 \frac{\ln^3 s}{1+s}ds.$$
Henceforce,
$$\lim_{x\to +\infty} \left( \frac{\ln^4 x}{4}-\int_0^x \frac{\ln^3 t}{1+t}dt \right)
=-2\int_0^1 \frac{\ln^3 t}{1+t}dt\ \left(=\frac{7\pi^4}{60}\right).$$

So our main goal is to calculate improper integral:
$$\color{blue}{\int_0^1 \frac{\ln^3 t}{1+t}dt\ \left(=-\frac{7\pi^4}{120}\right)}.$$

Here is a proof:
\begin{align*}
I&=\int_0^1\frac{-\ln^3t}{1+t}dt\cr
&=-\sum_{n=0}^\infty \int_{0}^1(-t)^n \ln^3 t\,dt\cr
&=-\sum_{k=0}^\infty \int_{0}^1(-t)^{2k} \ln^3 t\,dt-\sum_{k=0}^\infty \int_{0}^1(-t)^{2k+1} \ln^3 t\,dt\cr
&=\color{red}{-\sum_{k=0}^\infty \int_{0}^1t^{2k} \ln^3 t\,dt+\sum_{k=0}^\infty \int_{0}^1 t^{2k+1} \ln^3 t\,dt}\cr
&\color{red}{=\sum_{k=0}^\infty\frac{6}{(2k+1)^4}-\sum_{k=0}^\infty\frac{6}{(2k+2)^4}}\cr
&=\sum_{k=0}^\infty\frac{6}{(2k+1)^4}+\sum_{k=0}^\infty\frac{6}{(2k+2)^4}-2\sum_{k=0}^\infty\frac{6}{(2k+2)^4}\cr
&=6\zeta(4)-\frac{6}{8}\zeta(4)\cr
&=\frac{21}{4}\zeta(4)\cr
&=\frac{7\pi^4}{120}.
\end{align*}

Indeed, we use the following result (red part), the change of variables
$x=e^{-t}$ shows that
\begin{align*}\color{red}{
\int_0^1x^n\ln^p(1/x)\,dx}
&\color{red}{=\int_0^\infty e^{-(n+1)t}t^pdt}\\
&\color{red}{=\frac{1}{(n+1)^{p+1}}\int_0^\infty e^{-u}u^pdu}\\
&\color{red}{=\frac{\Gamma(p+1)}{(n+1)^{p+1}}.}
\end{align*}

A: By using Puiseux series I get that $\int_{0}^{x}\frac{\ln^3 t}{1+t} dt=(\frac{\ln^4 x}{4}-\frac{7\pi^4}{60})+O(\frac{1}{x})$
So it may be $\frac{7\pi^4}{60}$ at last.
