Calculate $\int \log(1+\log(x))x^ndx$ How to calculate the following integral?
$$\int \log(1+\log(x))x^ndx,$$
$n$ is an integer $\in N$.
 A: $\textbf{small hint}$
This was a direct result of the Mathematica output as shown by user* in the comments above. 
$$\int \log(1+\log(x))x^ndx = \int \log(\log(ex))x^ndx$$
substitute $t = ex$
we find
$$
\begin{eqnarray}
\int \log(1+\log(x))x^ndx &=& \frac{1}{e}\int \log(\log(t))\left(\frac{t}{e}\right)^ndt \\
&=& e^{-(n+1)}\int\log(\log(t))t^ndt
\end{eqnarray}
$$
make another sub $\log t = u$
we find
$$
e^{-(n+1)}\int \log u \mathrm{e}^{(n+1)u}du
$$
using the fact
$$
\int \mathrm{e}^{cx}\ln x = \frac{1}{c}\left(\mathrm{e}^{cx}\ln |x| - \mathrm{Ei}(cx)\right)
$$
we find that with $c=n+1$ and $x = u$
$$
e^{-(n+1)}\int \log u \mathrm{e}^{(n+1)u}du = e^{-(n+1)}\frac{1}{n+1}\left(\mathrm{e}^{(n+1)u}\ln |u| - \mathrm{Ei}[(n-1)u]\right)
$$
and you should be able to finish off with subbing in $u = 1+ \log(x)$
A: I think it can be solved by this way
$$\int \ln(1+\ln x)x^n dx$$
using integration by parts
$$u=\ln(1+\ln x)\Rightarrow
du=\frac{\frac{1}{x}}{1+\ln x}dx=\frac{dx}{x(1+\ln x)}\\
dv=x^ndx\Rightarrow v=\frac{x^{n+1}}{n+1}\\
\int \ln(1+\ln x)x^n dx=\ln(1+\ln x)\frac{x^{n+1}}{n+1}-\frac{1}{n+1}\int\frac{x^n}{1+\ln x}dx=\\
\frac{1}{n+1}\left[\ln(1+\ln x)x^{n+1}-\int\frac{x^n}{1+\ln x}dx\right]=\\
\frac{1}{n+1}\left[\ln(1+\ln x)x^{n+1}-\int\frac{x^n}{\ln ex}dx\right]$$
then
$$\int\frac{x^n}{\ln ex}dx\\
u=\ln ex\\
e^u=ex\Rightarrow x=e^{u-1}\\
du=\frac{dx}{x}\Rightarrow dx=xdu=e^{u-1}du\\
\int\frac{x^n}{\ln ex}dx=\int\frac{(e^{u-1})^n}{u}e^{u-1}du=\int\frac{e^{nu-n}e^{u-1}}{u}du=\\
\int\frac{e^{(n+1)u}e^{-(n+1)}}{u}du=e^{-(n+1)}\int\frac{e^{(n+1)u}}{u}du=\\
e^{-(n+1)}\text{Ei}[(n+1)u]=e^{-(n+1)}\text{Ei}[(n+1)\ln ex]=e^{-(n+1)}\text{Ei}[(n+1)(1+\ln x)]$$
which gives
$$\frac{\ln(1+\ln x)x^{n+1}-e^{-(n+1)}\text{Ei}[(n+1)(1+\ln x)]}{n+1}$$
