I’m trying to find $\int x^x \, dx$, but the only thing I know how to do is this:

Let $u=x^x$.

$$\begin{align}

\int x^x \, dx&=\int u \, du\\[6pt]

&=\frac{u^2}{2}\\[6pt]

&=\dfrac{\left(x^x\right)^2}{2}\\[6pt]

&=\frac{x^{2x}}{2}

\end{align}$$But it’s certain that this isn’t the correct way to evaluate that, and the answer must be wrong.

**Answer**

As noted in the comments, your derivation contains a mistake.

To answer the question, this function can not be integrated in terms of elementary functions. So there is no “simple” answer to your question, unless you are willing to consider a series approximation, obtained by expanding the exponential as a series:

$$\int{x^xdx} = \int{e^{\ln x^x}dx} = \int{\sum_{k=0}^{\infty}\frac{x^k\ln^k x}{k!}}dx$$

**Attribution***Source : Link , Question Author : Garmen1778 , Answer Author : QuantumDot*