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.

 A: Remember the following anagram:
Logarithmic
Inverse
Algebraic
Trigonometric
Exponential
This helps to choose what should be used as your $u$ (with a few exceptions). In your integrand, you have a $y$ (algebraic) and a logarithm $\ln(y)$ (logarithmic). So, you should set $u = \ln(y)$ and $\text{d}v = y\text{ d}y$. Hence, using integration by parts, you get 
$$\int y\ln(y)\text{ d}y = \ln(y)\cdot \dfrac{y^2}{2}-\int\dfrac{y^2}{2}\cdot \dfrac{1}{y}\text{ d}y = \dfrac{1}{2}y^2\ln(y)-\int\dfrac{1}{2}y\text{ d}y\text{.}$$
The last integral above is very straightforward.
See also https://math.berkeley.edu/~ehallman/math1B/IntByParts.pdf, http://www.phys.ttu.edu/~ritlg/courses/p4307/integration_by_parts/LIATEandTABULAR.pdf.
A: You integrated by parts with the wrong term.
Also consider letting $$x = ln(y)$$
So $dy = e^xdx$ and $\int y \ln(y)dy = \int e^{2x}xdx$
Which makes things easier to see.
A: 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:
$$\int_0^1 h(y)dy = -\frac{1}{4}.$$
A: By parts:
$$ \int y \ln y dy = \frac{1}{2} \int \ln y d(y^2) = \frac{ y^2 \ln y }{2} - \frac{1}{2} \int y^2 \frac{dy}{y} = \frac{4y^2 \ln y - y^2}{4} + C$$
A: $$\int_0^1y\ln ydy={y^2\over2}\ln y-\int_0^1{y^2\over2}{1\over y}dy=\left[{y^2\over2}\ln y\right]_0^1-\left[{y^2\over4}\right]_0^1=\left[{y^2\over2}\left(\ln y-{1\over2}\right)\right]_0^1=\\=\lim_{\epsilon\to0}\left[{y^2\over2}\left(\ln y-{1\over2}\right)\right]_{\epsilon}^1=-{1\over4}$$
A: You have $\int_0^1 y\ln(y)dy=y(y\ln(y)-y)|_0^1 -(\int_0^1 y\ln(y)dy-\int_0^1 ydy)$. Then just add $\int_0^1 y\ln(y)dy$ to both sides to get $2\int_0^1y\ln(y)dy=y(y\ln(y)-y)|_0^1+\int_0^1 ydy$. Factoring out that $2$ will get you your answer.
