9. $\\int y \\ln y dy$

Explanation:

Step1: Use integration by parts formula

The integration by parts formula is $\int u;dv=uv - \int v;du$. Let $u = \ln y$, then $du=\frac{1}{y}dy$. And let $dv=y;dy$, then $v=\frac{y^{2}}{2}$.

Step2: Substitute into the formula

Substitute $u$, $v$, $du$ and $dv$ into the integration by parts formula: [ \begin{align*} \int y\ln y;dy&=\frac{y^{2}}{2}\ln y-\int\frac{y^{2}}{2}\cdot\frac{1}{y}dy\ &=\frac{y^{2}}{2}\ln y-\frac{1}{2}\int y;dy \end{align*} ]

Step3: Integrate the remaining integral

Integrate $\int y;dy$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), when $n = 1$, $\int y;dy=\frac{y^{2}}{2}+C$. [ \begin{align*} \int y\ln y;dy&=\frac{y^{2}}{2}\ln y-\frac{1}{2}\cdot\frac{y^{2}}{2}+C\ &=\frac{y^{2}}{2}\ln y-\frac{y^{2}}{4}+C \end{align*} ]

Answer:

$\frac{y^{2}}{2}\ln y-\frac{y^{2}}{4}+C$