Question #6d541

Dec 15, 2017

${e}^{x} \left({x}^{2} + 1\right) + c$

Explanation:

Take the integral

$\int {e}^{x} {\left(x + 1\right)}^{2} \mathrm{dx}$

Expanding the integrand ${\left(x + 1\right)}^{2} = {x}^{2} + 2 x + 1$

$\int \left({e}^{x} {x}^{2} + 2 {e}^{x} x + {e}^{x}\right) \mathrm{dx}$

separate

$\int {e}^{x} {x}^{2} \mathrm{dx} + \int 2 {e}^{x} x \mathrm{dx} + \int {e}^{x} \mathrm{dx}$

For the integrand ${e}^{x} {x}^{2}$, integrate by parts,
$\int f \mathrm{dg} = f g - \int g \mathrm{df}$

$f = {x}^{2} , \mathrm{dg} = {e}^{x} \mathrm{dx}$
$\mathrm{df} = 2 x \mathrm{dx}$ $g = {e}^{x}$

${e}^{x} {x}^{2} - \int {e}^{x} 2 x \mathrm{dx}$

The integral of ${e}^{x} = {e}^{x} + c$

$\int {e}^{x} {x}^{2} \mathrm{dx} + \int 2 {e}^{x} x \mathrm{dx} + \int {e}^{x} \mathrm{dx} = {e}^{x} {x}^{2} - \int {e}^{x} 2 x \mathrm{dx} + \int 2 {e}^{x} x \mathrm{dx} + {e}^{x} + c$

$\int {e}^{x} {x}^{2} \mathrm{dx} + \int 2 {e}^{x} x \mathrm{dx} + \int {e}^{x} \mathrm{dx} = {e}^{x} {x}^{2} + {e}^{x} + c$

$\int {e}^{x} {x}^{2} \mathrm{dx} + \int 2 {e}^{x} x \mathrm{dx} + \int {e}^{x} \mathrm{dx} = {e}^{x} \left({x}^{2} + 1\right) + c$