How to solve $\int\ x^{\ln x} dx$? How to solve this integral


  $$\int\ x^{\ln x} dx$$


step by step?
We assume $x^{\ln x}$ is well defined.

Then we make the change of variable $x=e^{u}$, $dx=e^{u}\:du$, obtaining
$$
\begin{align}
\int x^{\ln x}\:dx&amp;=\int e^{(\ln x)^2}dx
\\&amp;=\int e^{u^2+u}du
\\&amp;=e^{-1/4}\int e^{(u+1/2)^2}du
\\&amp;=\frac{\sqrt{\pi }}2e^{-1/4}\: \text{erfi}\left(u+\tfrac12\right)
\end{align}
$$ where we have used the special function $\text{erfi}(\cdot)$.

Finally,


  $$
\int x^{\ln x}\:dx=\frac{\sqrt{\pi }}2e^{-1/4}\: \text{erfi}\left(\ln x+1/2\right)+C.
$$


One may notice that
$$
\operatorname{erfi}(z)= \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{z^{2n+1}}{n! (2n+1)},\quad z \in \mathbb{C}.
$$
get $$u=Lnx$$
so $$du=\frac{1}{x}dx$$
the main integral became
$$\int x^{u}xdu=\int x^{u+1}du\overset{x=e^{u}}{\rightarrow}\int (e^{u})^{u+1}du=\int e^{u^2+u}=\int e^{(u+\frac{1}{2})^{2}-\frac{1}{4}}du\overset{p=u+\frac{1}{2}}{\rightarrow}e^{\frac{-1}{4}}\int e^{p^2}dp$$
We can calculate the last integral by double integration.
$$I=\int e^{p^2}dp=\int e^{q^2}dq$$
so $$I^2=\int \int e^{p^2+q^2}dpdq\overset{r^{2}=p^2+q^2}{\rightarrow}\int \int re^{r^2}drd\theta =2\pi\frac{1}{2}e^{r^2}$$
finally $$I=\sqrt{\pi}e^{r}$$
and your answer is $$ans=\sqrt{\pi}e^{r-\frac{1}{4}}$$