2. $int y^{2}e^{2y}dy$

Explanation:

Step1: Apply integration - by - parts formula $\int u dv=uv-\int v du$

Let $u = y^{2}$, $dv=e^{2y}dy$. Then $du = 2y dy$, $v=\frac{1}{2}e^{2y}$. So, $\int y^{2}e^{2y}dy=\frac{1}{2}y^{2}e^{2y}-\int y e^{2y}dy$.

Step2: Apply integration - by - parts again on $\int y e^{2y}dy$

Let $u = y$, $dv=e^{2y}dy$. Then $du = dy$, $v=\frac{1}{2}e^{2y}$. So, $\int y e^{2y}dy=\frac{1}{2}ye^{2y}-\int\frac{1}{2}e^{2y}dy$.

Step3: Integrate $\int\frac{1}{2}e^{2y}dy$

$\int\frac{1}{2}e^{2y}dy=\frac{1}{4}e^{2y}+C$.

Step4: Substitute back

$\int y e^{2y}dy=\frac{1}{2}ye^{2y}-\frac{1}{4}e^{2y}+C$. $\int y^{2}e^{2y}dy=\frac{1}{2}y^{2}e^{2y}-\left(\frac{1}{2}ye^{2y}-\frac{1}{4}e^{2y}\right)+C=\frac{1}{2}y^{2}e^{2y}-\frac{1}{2}ye^{2y}+\frac{1}{4}e^{2y}+C$.

Answer:

$\frac{1}{2}y^{2}e^{2y}-\frac{1}{2}ye^{2y}+\frac{1}{4}e^{2y}+C$