<s>[INST] Help with integrating 101: $\int y \ln{y}\,\mathbb dy$ Would appreciate it if someone would please help me solve this 
$$\int y\;\ln y\, \mathbb dy$$
taking time to explain reason for each step taken. 
Thanks in advance!
 [/INST] Here is a method of doing this by substitution. Put $y=e^{x}$. Then you have $dy = e^{x} \ dx$.  Substituting we have 
\begin{align*}
\int e^{2x} \cdot x \ dx &= \frac{1}{4}\int e^{t} \cdot t\ dt \qquad \Bigl[ \text{substituting} \ t = 2x \Bigr] \\ &= \frac{1}{4}\int e^{t} \cdot \bigl( t + 1 \bigr) \ dt - \frac{1}{4}\int e^{t} \ dt \\ &=  \frac{1}{4}e^{t} \cdot t -\frac{1}{4} e^{t} + C \qquad\qquad\qquad \Bigl[ \because \small \int e^{x} \cdot \Bigl( f(x) + f'(x) \Bigr) \ dx = e^{x} \cdot f(x) + C \ \Bigr] \\ &=\frac{1}{4}\cdot e^{2x} \cdot 2x - \frac{1}{4}\cdot e^{2x} +C \\ &=\log{y} \cdot y^{2} \times \frac{1}{2} - \frac{1}{4} \cdot y^{2} +C
\end{align*}
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