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.
 A: This should be a comment, but it would be too long for the comments section, so I'll write it here. Please do not vote up / down unless you find I wrote a real malarkey. In that case, I'll provide to delete this or to improve it.
As we said in the comments, there are no expressions in terms of simple and elementary functions, and also most probably there are no answers in terms of any functions, as @Claude Leibovici stated.
The other problem is that your integrals are undefined, so either numerical approaches are available. I'll provide to show you some substitutions to make things clearer. 
1
In this integral you might naively try the substitution
$$e^{x^2} = t ~~~~~ x^2 = \ln(t) ~~~~~ x = \sqrt{\ln(t)} ~~~~~ \text{d}t = 2t\ \sqrt{\ln(t)}$$
gaining
$$\frac{1}{2}\int\ \frac{(\ln(t))^{t/2 - 1/2}}{t}\ \text{d}t$$
and since it involves the form $t^{-1}[F(t)]^t$ the possibilities it can be integrated are like zero. I don't think one may find a more suitable expression (despite the fact it makes me to think about the Logarithmic Integral Special Function, but it's more complex since it has a $t$ in the exponent itself).
2
For this one, it could be good to use (for $t > 0$)
$$x = e^t ~~~~~~~ \text{d}x = e^t\ \text{d}t$$
to get
$$\int\frac{e^t}{e^t + t^2}\ \text{d}t$$
We may collect the $e^t$ factor to have
$$\int\frac{\text{d}t}{1 + t^2e^{-t}}$$
Which could be treated in several ways, if the integral had extrema. Having not, we cannot say that much.
Notice also that this substitution works for $t > 0$ only, so other possible ways are opened.
A: 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)
$=\ln x+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^{2n}\dfrac{(-1)^{n+1}(2n)!\ln^kx}{n^{2n-k+1}k!x^n}+C$
