Answer
$$\int t^2e^{4t}dt=\frac{e^{4t}}{32}(8t^2-4t+1)+C$$
Work Step by Step
$$A=\int t^2e^{4t}dt$$
Set $u=t^2$ and $dv=e^{4t}dt$
Then we have $du=2tdt$ and $v=\frac{1}{4}e^{4t}$
Using the formula $\int udv= uv-\int vdu$:
$$A=\frac{t^2e^{4t}}{4}-\int\frac{1}{4}e^{4t}\times2tdt$$ $$A=\frac{t^2e^{4t}}{4}-\frac{1}{2}\int te^{4t}dt$$
Set $u=t$ and $dv=e^{4t}dt$
Then we have $du=dt$ and $v=\frac{1}{4}e^{4t}$
Using the formula $\int udv= uv-\int vdu$:
$$A=\frac{t^2e^{4t}}{4}-\frac{1}{2}\Big(\frac{te^{4t}}{4}-\frac{1}{4}\int e^{4t}dt\Big)$$ $$A=\frac{t^2e^{4t}}{4}-\frac{te^{4t}}{8}+\frac{1}{8}\int e^{4t}+C$$ $$A=\frac{t^2e^{4t}}{4}-\frac{te^{4t}}{8}+\frac{e^{4t}}{32}+C$$ $$A=\frac{e^{4t}}{32}(8t^2-4t+1)+C$$
