<s>[INST] Integrate $\int y \ln y \; \mathrm{d}y$ Can you please show me how to integrate this?

$$\int y \ln y \;  \mathrm{d}y$$

I attempted integration by parts but it is getting more complicated.

 [/INST] Do you know integration by parts?
$$\int fg' = fg - \int f' g.$$
Let's pose $f = \log(y)$ and $g' = y$. Then:
$$f' = \frac{1}{y}, g = \frac{1}{2}y^2.$$
Joining all:
$$\int y\log(y) dy = \frac{1}{2}y^2\log(y) - \int \frac{1}{y}\frac{1}{2}y^2 dy = \\
\frac{1}{2}y^2\log(y) - \frac{1}{4}y^2 + c. $$
If $h(y) = y\log(y)$, then let's call $H(y) =  \frac{1}{2}y^2\log(y) - \frac{1}{4}y^2.$
Beware of $0$!!!
Indeed, the function $y \log(y)$ is not defined for $y=0$ (what happens to logarithm when $y$ goes to $0$?). Then:
$$\int_0^1 h(y)dy = H(1) - \lim_{y \to 0} H(y).$$
The first part is simple:
$$H(1) = \frac{1}{2}1^2 \cdot \log(1) - \frac{1}{4}1^2 = -\frac{1}{4}.$$
The second part is "harder":
$$\lim_{y \to 0} H(y) = \lim_{y \to 0}\left(  \frac{1}{2}y^2\log(y) - \frac{1}{4}y^2\right) = \lim_{y \to 0}\left(  \frac{1}{2}y^2\log(y)\right) - 0 = \\
= 0 - 0 = 0.$$
Then:
$$\int_0^1 h(y)dy = -\frac{1}{4}.$$
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