Series Expansion Of An Integral. I want to find the first 6 terms for the series expansion of this integral:

$$\int x^x~dx$$

My idea was to let: $$x^x=e^{x\ln x}$$

From that we have: $$\int e^{x\ln x}~dx$$

The series expansion of $e^x$ is: $$\sum\limits_{n=0}^\infty\frac{x^n}{n!}$$

Then we have:

$$\int e^{x\ln x}~dx=\int\sum\limits_{n=0}^\infty\frac{(x\ln x)^n}{n!}dx$$

I integrated the first 3 terms (since it was the easiest the work with):
$$x+\frac{1}{4}x^2(2\ln x-1)+\frac{1}{54}x^3(9(\ln x)^2-6\ln x+2)+\cdots$$

Is there any other efficient way of getting the first 6 terms of the series expansion of the integral?
You are in the right approach. Then you have to face $\int x^n(\ln x)^n~dx$ for non-negative integers $n$ :

$\int x^n(\ln x)^n~dx$

$=\int(\ln x)^n~d\left(\dfrac{x^{n+1}}{n+1}\right)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{x^{n+1}}{n+1}d((\ln x)^n)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)x}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{nx^n(\ln x)^{n-1}}{n+1}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\int\dfrac{n(\ln x)^{n-1}}{n+1}d\left(\dfrac{x^{n+1}}{n+1}\right)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{x^{n+1}}{n+1}d\left(\dfrac{n(\ln x)^{n-1}}{n+1}\right)$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^{n+1}(\ln x)^{n-2}}{(n+1)^2x}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\int\dfrac{n(n-1)x^n(\ln x)^{n-2}}{(n+1)^2}dx$

$=\cdots\cdots$

$\vdots$

$\vdots$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}-\int\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times1)x^n}{(n+1)^n}dx$

$=\dfrac{x^{n+1}(\ln x)^n}{n+1}-\dfrac{nx^{n+1}(\ln x)^{n-1}}{(n+1)^2}+\cdots\cdots+\dfrac{(-1)^{n-1}(n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^n}+\dfrac{(-1)^n(n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+1}}+C$

$=\dfrac{(n+1)x^{n+1}(\ln x)^n}{(n+1)^2}-\dfrac{(n+1)nx^{n+1}(\ln x)^{n-1}}{(n+1)^3}+\cdots\cdots+\dfrac{(-1)^{n-1}((n+1)n(n-1)\cdots\cdots\times2)x^{n+1}\ln x}{(n+1)^{n+1}}+\dfrac{(-1)^n((n+1)n(n-1)\cdots\cdots\times1)x^{n+1}}{(n+1)^{n+2}}+C$

$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}(n+1)!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+2}}+C$

$=\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$

$\therefore\int\sum\limits_{k=0}^n\dfrac{(x\ln x)}{n!}dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$

Hence $\int x^x~dx=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}x^{n+1}(\ln x)^k}{k!(n+1)^{n-k+1}}+C$
Try taking a Taylor series for $\log x$, multiplying it by $x$, and plugging the result into the Taylor series for $e^x$. You can easily compute the first few terms.

Your choice of series for $\log x$ will affect where the result converges, so nothing will be valid everywhere. If this were a definite integral, it might be better.