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!
 A: We go for integration by parts, since we are looking at the product of two unrelated functions, and there is no obvious substitution.
I do not know what notation you use when setting up an integration by parts, so I will guess. The integration by parts "formula" can be written in two completely equivalent ways, apart from the choice of names. These are
$$\int u\,dv=uv -\int v\,du\qquad\text{and}$$
$$\int v\,du=uv -\int u\,dv.$$
This formula comes directly from the Product Rule for derivatives. Please consult your text for details.  In integration by parts, often the key question is: what shall I choose as $u$ (and what shall I choose as $dv$).
Maybe we should let $u=y$. Then $dv$ must be $\ln y \,dy$.  That sounds good, we will then have $du=dy$, which is a simplification.  Great.  We have let $dv=\ln y\,dy$, so we need to find $v$, that is, to integrate $\ln y$.  Looks a little ugly, or at least not immediate.  
Let's not despair, maybe the integral of $\ln y$ will not be too hard.  But we might as well take a side glance at the other plausible choice,  $du=y\,dy$, $v=\ln y$.  Then $u=y^2/2$ (we don't worry about constants of integration yet). Ouch, looks worse than $y$! And what looks worse usually is worse. But let's persist: $dv=(1/y)\,dy$.  That's really good, ugly stuff is gone, we are basically finished.  
So, formally now, let $du=y\,dy$, let $v=\ln y$.  Then $u=y^2/2$, $dv=(1/y)\,dy$.
(Do lay it out more nicely than I have.) So we have
$$\int y\ln y \,dy=uv-\int u\,dv=\frac{y^2}{2}\ln y-\int \frac{y}{2}dy.$$
Integrate, remembering now about the constant of integration.
$$\int y\ln y\,dy=\frac{y^2\ln y}{2}-\frac{y^2}{4}+C.$$
Comment: When we integrate $x^3e^{2x}$, for example, the integration by parts strategy focuses on lowering the power of $x$.  That is "usually" a good strategy.  Exceptions to this general heuristic are things like $x^3\ln x$, and $x^3 \arctan x$.  
Extreme examples are $\int \ln x\,dx$, $\int \arctan x\,dx$, and $\int \arcsin x\, dx$, where it looks as if we don't even have a "product".  But if, for example in $\int \ln x \,dx$, we let $du=dx$, $v=\ln x$, everything works out quickly.   
Added: Could substitution be a feasible approach?  It seems sensible to try for $z=\ln y$, or equivalently $y=e^z$.  Then $dy=e^z\,dz$.  Our integral becomes
$$\int ze^{2z}\,dz.$$
This is an integration by parts, so the substitution didn't buy us much!  But it's a more "normal" integration by parts.  The rest is routine.
A: 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*}
A: Here's another way to arrive at the result (which also provides a partial intuition).
Fix $0 < y \leq 1$. Then,
$$
\int_0^y {u\ln u \,du}  = \int_0^y {u\bigg( - \int_u^1 {\frac{1}{v}} \,dv\bigg)du}  = \int_{u = 0}^y {\bigg(\int_{v = u}^1 {\frac{{ - u}}{v}\,dv\bigg)du} } .
$$
Interchanging the order of integration (as usual, it is helpful to draw a picture) yields
$$
\int_0^y {u\ln u \,du}  =  \int_{v = 0}^y {\bigg(\int_{u = 0}^v {\frac{{ - u}}{v}\,du\bigg)dv} } +
\int_{v = y}^1 {\bigg(\int_{u = 0}^y {\frac{{ - u}}{v}\,du} \bigg)dv} .
$$
Hence
$$
\int_0^y {u\ln u \,du}  = \int_{v = 0}^y {\frac{1}{v}} \frac{{ - v^2 }}{2}\,dv + \int_{v = y}^1 {\frac{1}{v}\frac{{ - y^2 }}{2}\,dv}  =  - \frac{{y^2 }}{4} + \frac{{y^2 \ln y}}{2}.
$$
A: Use integration by parts:
$$ \int \frac{df}{dy}\  g \ \ \text{d}y = fg - \int f \ \frac{dg}{dy} \ \ \text{d}y $$
Thus we get
$$ \int \frac{d (y^2/2)}{dy}\ \ln y \ \ \text{d}y = \frac{y^2}{2}\ \ln y - \int \frac{y^2}{2} \frac{d \ln y}{dy} \ \ \text{d}y$$
Now $\displaystyle \frac{d \ln y}{dy} = \frac{1}{y}$.
The last integral becomes
$$\int \frac{y^2}{2} \frac{1}{y} \ \ \text{d}y = \int \frac{y}{2} \ \ \text{d}y = \frac{y^2}{4}$$
Thus your integral is
$$ \frac{y^2 \ln y}{2} - \frac{y^2}{4} + C$$
The reason to use integration by parts is that the derivative of $\displaystyle \ln y$ is $\displaystyle \frac{1}{y}$ which cancels out with one of the $y$ got from the other term.
For instance try the same technique with the following:
$$\int y^{1000} \ln y \ \text{d}y$$
