Prove that $\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\cdots $ Prove that 

$$\int \frac{x^m dx}{\ln x}=\ln(\ln x)+\frac{(m+1)\ln x}{1}+\frac{(m+1)^2 \ln ^2 x}{1 \times 2^2}+\frac{(m+1)^3 \ln^3 x}{1 \times 2 \times 3^2}+\cdots \infty $$

My Try:
I started with $$I=\int \frac{x^m}{\ln x}=\int \frac{x^{m+1} dx}{x \ln x}$$
Now by using parts taking $u=x^{m+1}$ and $v=\frac{1}{x \ln x}$ we get
$$I=x^{m+1} \times \ln(\ln x)-(m+1)\int x^m \ln (\ln x)dx$$ 
can i have any clue to proceed
 A: With substitution $x=e^u$:
\begin{align}
\int \frac{x^m}{\ln x}dx
&=\int\dfrac{1}{u}e^{u(m+1)}du \\
&=\int\dfrac{1}{u}\sum_{n=0}^\infty \dfrac{u^n(m+1)^n}{n!}du \\
&=\int\left(\dfrac{1}{u}+\sum_{n=1}^\infty \dfrac{u^{n-1}(m+1)^n}{n!}du\right) \\
&=\ln u+\sum_{n=1}^\infty \dfrac{u^n(m+1)^n}{n.n!} \\
&=\ln\ln x+\sum_{n=1}^\infty \dfrac{\ln^nx(m+1)^n}{n.n!}
\end{align}
A: Since differentiating is usually easier than finding primitives, and you are given a candidate, the simplest way is to differentiate the right hand side. Write that as
$$f(x) = \ln (\ln x) + \sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^n}{n\cdot n!}.\tag{1}$$
The series converges locally uniformly on $(0,+\infty)$, so it can be differentiated term-by-term. This yields
\begin{align}
f'(x) &= \frac{1}{x\ln x} + \sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^{n-1}}{n!\cdot x} \\
&= \frac{1}{x\ln x} + \frac{1}{x\ln x}\sum_{n = 1}^{\infty} \frac{(m+1)^n(\ln x)^n}{n!} \\
&= \frac{1}{x\ln x} \sum_{n = 0}^{\infty} \frac{(m+1)^n(\ln x)^n}{n!} \\
&= \frac{1}{x\ln x} e^{(m+1)\ln x} \\
&= \frac{x^{m+1}}{x\ln x} \\
&= \frac{x^m}{\ln x}
\end{align}
for $x > 1$, so the right hand side is indeed a primitive of the integrand.
