Compute $\int_1^e \frac{dx}{x(x+(\ln x)^2)}$ My friend asked me how to integrate the following:
$$\int_1^e \frac{dx}{x(x+(\ln x)^2)}$$

How am I going to solve this?Any help is greatly appreciated.
Thanks.
Assume $u=\ln x$, then
\begin{align}
I:=\int_1^e \frac{dx}{x(x+(\ln x)^2)}
=\int_0^1 \frac{e^u du}{e^u(e^u+u^2)}
&amp;=\int_0^1 \frac{ du}{e^u+u^2}
\\
&amp;=\int_0^1 \frac{e^{-u} du}{1+u^2e^{-u}}
\end{align}
Using geometric series $$\frac{1}{1+u^2e^{-u}}=\sum_{n=0}^{\infty}{(-1)^nu^{2n} e^{-nu} }$$ term by term integration, we have
\begin{align}
I=\sum(-1)^n (\int_0^1 {  u^{2n} e^{-(n+1)u}du}=J_n)
\end{align}
Integrating by parts we get (from tables) $$J_n=-e^{-(n+1) u}\sum_{k=0}^{2n}\frac{(2n)!}{(2n-k)!}(n+1)^{-(k+1)}u^{2n-k}$$
Assume $u=\ln x$, then
\begin{align}
I:=\int_1^e \frac{dx}{x(x+(\ln x)^2)}
=\int_0^1 \frac{e^u du}{e^u(e^u+u^2)}
&amp;=\int_0^1 \frac{ du}{e^u+u^2}
\\
&amp;=\int_0^1 \frac{e^{-u} du}{1+u^2e^{-u}}
\end{align}
Using geometric series $$\frac{1}{1+u^2e^{-u}}=\sum_{n=0}^{\infty}{(-1)^nu^{2n} e^{-nu} }$$ term by term integration, we have
\begin{align}
I=\sum(-1)^n (\int_0^1 {  u^{2n} e^{-(n+1)u}du}=J_n)
\end{align}
Integrating by parts we get (from tables) $$J_n=-e^{-(n+1) u}\sum_{k=0}^{2n}\frac{(2n)!}{(2n-k)!}(n+1)^{-(k+1)}u^{2n-k}$$