Question

Evaluate $\int x^2{e}^{x}dx$.

Answer 1

$\Rightarrow$ Let $u=x^2$ and $v=e^x$, then $du=2xdx$

Now i ntegration by parts states that

$\int u\left(x\right)v^{\prime }\left(x\right)dx=u\left(x\right)v\left(x\right)-\int v\left(x\right)u^{\prime }\left(x\right)dx$

Hence, $\int x^2{e}^{x}dx=x^2{e}^x-\int e^x\times 2xdx$

$\int x^2{e}^{x}dx=x^2{e}^x-2\int x{e}^{x}dx+C$ ------- ( 1 )

Now, we set $u=x,$ then $du=dx$

and $\int x{e}^{x}dx=x{e}^x-\int e^x\times 1\times dx$ or

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x$

Putting this in ( 1 ), we get

$\int x^2{e}^{x}dx=x^2{e}^x-2(x{e}^x-e^x)+C$

$\int x^2{e}^{x}dx=e^x(x^2-2x+2)+C$