how to solve $\int_0^1 x^4\left(\ln\frac{1}{x}\right)^3dx$? Solve the following integral  $$\int_0^1 x^4\left(\ln\frac{1}{x}\right)^3dx$$
my attempt:
$$\int_0^1 x^4\left(\ln\frac{1}{x}\right)^3dx$$ $$=-\int_0^1 x^4\left(\ln x\right)^3dx$$
substitute $ln x=t, \ dx=e^tdt$
$$=-\int_{-\infty}^0 (e^t)^4t^3e^t\ dt$$
$$=-\int_{-\infty}^0 t^3e^{5t}\ dt$$
$$=\int_0^{-\infty} t^3e^{5t}\ dt$$
I got stuck here. I am not sure how to proceed. please help me solve it.
Thanks.
 A: by using ${\int}\mathtt{f}\mathtt{g}' = \mathtt{f}\mathtt{g} - {\int}\mathtt{f}'\mathtt{g}
,f=(ln)^3,g=x^4$
$$=\int_0^1 x^4\left(\ln x\right)^3dx=\dfrac{x^5\ln^3\left(x\right)}{5}-{\displaystyle\int}\dfrac{3x^4\ln^2\left(x\right)}{5}\,\mathrm{d}x
$$ using integration by parts
$${\displaystyle\int}\dfrac{3x^4\ln^2\left(x\right)}{5}\,\mathrm{d}x
=\frac 35 (=\dfrac{x^5\ln^2\left(x\right)}{5}-{\displaystyle\int}\dfrac{2x^4\ln\left(x\right)}{5}\,\mathrm{d}x
)$$ and once again
$${\displaystyle\int}x^4\ln^3\left(x\right)\,\mathrm{d}x
=\\
=\dfrac{x^5\ln^3\left(x\right)}{5}-\dfrac{3x^5\ln^2\left(x\right)}{25}+\dfrac{6x^5\ln\left(x\right)}{125}-\dfrac{6x^5}{625}
$$
A: You are almost done.

*

*Using Gamma Function
Substitute $5t=-y\implies dt=-\frac{dy}{5}$
$$\begin{align*}
\int_0^{-\infty} t^3e^{5t}\ dt
&=\int_0^{\infty} \left(-\frac y5\right)^3e^{-y}\ \left(-\frac{dy}{5}\right)\\
&=\frac{1}{625}\int_0^{\infty} e^{-y}y^3dy\quad (\text{Recall Gamma \\Function})\\
&=\frac{1}{625}\Gamma{4}\\
&=\frac{6}{625}\\
\end{align*}$$

*

*Using Laplace Transform
Substitute $5t=-y\implies dt=-\frac{dy}{5}$
$$\begin{align*}
\int_0^{-\infty} t^3e^{5t}\ dt
&$=\int_0^{\infty} \left(-\frac y5\right)^3e^{-y}\ \left(-\frac{dy}{5}\right)\\
&=\frac{1}{625}\int_0^{\infty} e^{-y}y^3dy\quad (\text{Recall Laplace Transform})\\
&=\frac{1}{625}L\left[t^3\right]_{s=1}\quad \left(\because L[t^n]=\int_0^{\infty}e^{-st}t^n\ dt=\frac{\Gamma(n+1)}{s^{n+1}}\right)\\
&=\frac{1}{625}\left[\frac{\Gamma{4}}{s^4}\right]_{s=1}\\
&=\frac{\Gamma{4}}{625}\\
&=\frac{6}{625}\\
\end{align*}$$
