Please show your working.
7 Answers
I jus tried..!!!
Write x power x as 1.xx which is equal to tan(pi4)
apply ILATE....is the answer xx+1\2
I think the answer is xx(1+lnx).
y=xx.
logy=xlogx
differentiate
1ydydx=x1x+logx
You can figure out the rest.
I Think Something Like in this way..........
\hspace{-16}\mathbf{\int x^xdx=\int e^{x.\ln(x)}dx}$\\\\\\ $\mathbf{\int \left(1+\frac{x.\ln(x)}{1!}+\frac{x^2\ln^2(x)}{2!}+............\right)dx}$\\\\\\ $\mathbf{\int \sum_{n=1}^{\infty}\frac{x^n.\ln^n(x)}{n!}dx}$\\\\\\ $\mathbf{\sum_{n=1}^{\infty}\frac{1}{n!}.\int x^n.\ln^n(x)dx }$
Well I have an approach;
Integrate xx.1 by parts taking xx as Ist part and 1 as 2nd term (Is this correct?)
Then probably problem becomes simple!
Check out 63rd question in Arihant Integral Calculus in Exercise of Indefinite Integral! A similar problem!