Stuck on how to approach integral $\int_0^1x^n\log(x)\,dx$ I am being asked the following:
For $n \gt 0$, and by considering an appropriate limit, find:
$$\int_0^1x^n\log(x)\,dx$$
I want to try integration by parts, but I am confused about what the question means when it states by considering an appropriate limit.
 A: Note $\int_0^1x^n\,dx= \frac1{n+1}$ and
$$ \int_0^1x^n\log xdx= \frac {d}{dn } \int_0^1x^n\,dx =-\frac1{(n+1)^2}$$
A: 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
$$\int_0^1 x^n \ln(x)\mathrm{d}x=1^{n+1}\left(\frac{\ln(1)}{n+1}-\frac{1}{(n+1)^2}\right)=\frac{-1}{(n+1)^2}.$$
A: $$\int_0^1x^n\log(x)dx=\left[\log(x)\frac{x^{n+1}}{n+1}\right]_0^1-\int_0^1\frac{x^n}{n+1}dx=-\left[\frac{x^{n+1}}{(n+1)^2}\right]_0^1=-\frac{1}{(n+1)^2}$$
A: Using integration by parts:
Let $\displaystyle u=\ln(x) \Leftrightarrow u'=\frac{1}{x}$ and $\displaystyle v=\frac{x^{n+1}}{n+1}\Leftrightarrow v'=x^n$
\begin{align}
\mathcal{I}&=\int_0^1 x^n\ln(x)\mathrm{d}x\\
&=\frac{x^n\ln(x)}{n+1}\bigg\vert_0^1-\int_0^1\frac{x^{n+1}}{(n+1)x}\mathrm{d}x\\
&=-\int_0^1\frac{x^n}{n+1}\mathrm{d}x\\
&=-\frac{1}{n+1}\bigg(\frac{x^{n+1}}{n+1}\bigg\vert_0^1\bigg)\\
&=-\frac{1}{(n+1)^2}
\end{align}
Hence our final answer is:
$$\mathcal{I}=-\frac{1}{(n+1)^2}$$
Good luck!
A: Let $x=e^u$ and $dx=e^udu$
Then,
$$\int_{0}^{1} {x^nlnxdx}=\int_{-\infty}^{0} {ue^{un}e^udu}$$
$$=\int_{-\infty}^{0}ue^{(n+1)u}du$$
$$=\bigg[{{(n+1)u-1}\over{(n+1)^2}} {e^{(n+1)u}} ] _{-\infty}^{0}$$
$$=-{1\over {(n+1)^2}}$$
