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
