Stuck on how to approach integral $\int_0^1x^n\log(x)\,dx$

Note $\int_0^1x^n\,dx= \frac1{n+1}$ and

$$ \int_0^1x^n\log xdx= \frac {d}{dn } \int_0^1x^n\,dx =-\frac1{(n+1)^2}$$

When approaching limits using integration by parts, you need to ask yourself: "which parts of this integral will become simpler when I integrate or differentiate it?" So let's have a look at $$\int x^n \ln(x)\mathrm{d}x$$ Well, $x^n$ doesn't become much more complicated when you differentiate or integrate it. However, while $\ln(x)$ is difficult to integrate, its derivative is quite simple, namely, $1/x$. So we choose $u=\ln(x)$, $\mathrm{d}v=x^n\mathrm{d}x$, and from this we conclude $\mathrm{d}u=\frac{1}{x}\mathrm{d}x$, $v=\frac{x^{n+1}}{n+1}$. So, $$\int x^n \ln(x)\mathrm{d}x=\int u\mathrm{d}v=uv-\int v\mathrm{d}u=\ln(x)\frac{x^{n+1}}{n+1}-\int \frac{x^{n+1}}{n+1} \frac{1}{x}\mathrm{d}x$$ $$=x^{n+1}\left(\frac{\ln(x)}{n+1}-\frac{1}{(n+1)^2}\right)+C$$ In the limiting case of $n\to 0$ this gives the nice expression $$\int \ln(x)\mathrm{d}x=x\ln(x)-x+C$$

EDIT: This also shows that $$\int_0^1 x^n \ln(x)\mathrm{d}x=1^{n+1}\left(\frac{\ln(1)}{n+1}-\frac{1}{(n+1)^2}\right)=\frac{-1}{(n+1)^2}.$$