Weird indefinite integral homework questions I'm solving a couple of integration problems using the method of changing variables, and would like assistance with two particular problems that I can't seem to solve. I completed rest of the problems in this problem set without much effort, but these two seem impossible.
I've tried changing a few different variables in both problems, and I tried to calculate the solution with Wolfram Alpha, but neither of those had any avail.
$$\int x^{e^{x^2}}~dx$$
and
$$\int\frac{dx}{x+\ln^2x}$$
are the problems that I'm trying to solve. Any help is much appreciated.

For $\int x^{e^{x^2}}~dx$ ,
$\int x^{e^{x^2}}~dx$
$=\int e^{e^{x^2}\ln x}~dx$
$=\int\left(1+\sum\limits_{m=1}^\infty\dfrac{e^{mx^2}\ln^mx}{m!}\right)~dx$
Let $u=\ln x$ ,
Then $x=e^u$
$dx=e^u~du$
$\therefore\int\left(1+\sum\limits_{m=1}^\infty\dfrac{e^{mx^2}\ln^mx}{m!}\right)~dx$
$=x+\int\sum\limits_{m=1}^\infty\dfrac{u^me^ue^{me^{2u}}}{m!}~du$
$=x+\int\sum\limits_{m=1}^\infty\sum\limits_{n=0}^\infty\dfrac{m^nu^me^{(2n+1)u}}{m!n!}~du$
$=x+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^m\dfrac{(-1)^{m-k}m^nu^ke^{(2n+1)u}}{(2n+1)^{m-k+1}n!k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$=x+\sum\limits_{m=1}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^m\dfrac{(-1)^{m-k}m^nx^{2n+1}\ln^kx}{(2n+1)^{m-k+1}n!k!}+C$
For $\int\dfrac{dx}{x+\ln^2x}$ ,
Let $u=\ln x$ ,
Then $x=e^u$
$dx=e^u~du$
$\therefore\int\dfrac{dx}{x+\ln^2x}=\int\dfrac{e^u}{e^u+u^2}~du$
Case $1$: $u^2e^{-u}\leq1$ , i.e. $\dfrac{\ln^2x}{x}\leq1$
Then $\int\dfrac{e^u}{e^u+u^2}~du$
$=\int\dfrac{1}{1+u^2e^{-u}}~du$
$=\left(1+\int\sum\limits_{n=1}^\infty(-1)^nu^{2n}e^{-nu}\right)~du$
$=u+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n+1}(2n)!u^ke^{-nu}}{n^{2n-k+1}k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$=\ln x+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n+1}(2n)!\ln^kx}{n^{2n-k+1}k!x^n}+C$
