Integrate $y=\int x^x(\ln x+1)\ dx$

Question

$$y=\int x^x(\ln x+1)\ dx$$ I tried integration by parts but it became more complicated. So I did this: $$dy=x^x(\ln x+1)$$ $$\frac1{x^x}dy=\ln x+1$$ $$\int\frac1{x^x}\ dy=\int(\ln x+1)\ dx=x\ln x=\ln x^x$$ Comparing with $\int\frac1y\ dy=\ln y$, I concluded $y=x^x+K$. Is there anything wrong with this method?

Answer 1

Integration by parts is unnecessary, since the integral is amenable to a single, judiciously chosen substitution. Note $$x^x = e^{x \log x}.$$ Then, what is the derivative of $x \log x$ with respect to $x$?

Answer 2

First of all the $\int\frac1{x^x}\ dy$ is wrong step because $x$ and $y$ are two interdependent variables ( None of them is constant with respect to each other ).

Now for answer of $y=\int x^x(\ln x+1)\ dx$,

consider $$ \frac{d(x^x)}{dx}=\frac{d(e^{xlnx})}{dx}=e^{xlnx}\frac{d(x\ln{x})}{dx}=e^{xlnx}(\ln{x}+1)=x^x(\ln{x}+1) $$ So, $$ y=\int x^x(\ln x+1)\ dx=\int \frac{d(x^x)}{dx} dx=x^x+K $$

Answer 3

Put $x^x=y$ and find the $y'$. We have $$y=x^x/\ln$$ $$\ln y=\ln x^x/'$$ $$\ln y=x\ln x$$ $$y'\cdot\frac{1}{y}=\ln x+1/\cdot y$$ $$y'=y(\ln x+1)$$ $$y'=x^x(\ln x+1)$$

For the given example, we have:

$$\int x^x(\ln x+1)dx=\int y' dy=\left(\int ydy\right)'=y=x^x+C$$