# How do you integrate int x^3 e^x dx  using integration by parts?

Mar 1, 2016

$\setminus \int {x}^{3} {e}^{x} \mathrm{dx} = {x}^{3} {e}^{x} - 3 \left({x}^{2} {e}^{x} - 2 \left({e}^{x} x - {e}^{x}\right)\right) + C$

#### Explanation:

$\setminus \int {x}^{3} {e}^{x} \mathrm{dx}$

Applying integration by parts as: $\setminus \int u v ' = u v - \setminus \int u ' v$
$u = {x}^{3} , u ' = 3 {x}^{2} , v ' = {e}^{x} , v = {e}^{x}$

$= {x}^{3} {e}^{x} - \setminus \int 3 {x}^{2} {e}^{x} \mathrm{dx}$

$\setminus \int 3 {x}^{2} {e}^{x} \mathrm{dx} = 3 \left({x}^{2} {e}^{x} - 2 \left({e}^{x} x - {e}^{x}\right)\right)$ as under

=$\setminus \int 3 {x}^{2} {e}^{x} \mathrm{dx}$

taking the constant out as: $\setminus \int a \setminus \cdot f \left(x\right) \mathrm{dx} = a \setminus \cdot \setminus \int f \left(x\right) \mathrm{dx}$

$= 3 \setminus \int {x}^{2} {e}^{x} \mathrm{dx}$

Applying integration by parts as: $\setminus \int u v ' = u v - \setminus \int u ' v$

$u = {x}^{2} , u ' = 2 x , v ' = {e}^{x} , v = {e}^{x}$

$= 3 \left({x}^{2} {e}^{x} - \setminus \int 2 x {e}^{x} \mathrm{dx}\right)$

taking the constant out as: $\setminus \int a \setminus \cdot f \left(x\right) \mathrm{dx} = a \setminus \cdot \setminus \int f \left(x\right) \mathrm{dx}$

$= 3 \left({x}^{2} {e}^{x} - 2 \setminus \int x {e}^{x} \mathrm{dx}\right)$

Applying integration by parts as: $\setminus \int u v ' = u v - \setminus \int u ' v$

$u = x , u ' = 1 , v ' = {e}^{x} , v = {e}^{x}$

$= 3 \left({x}^{2} {e}^{x} - 2 \left(x {e}^{x} - \setminus \int 1 {e}^{x} \mathrm{dx}\right)\right)$

$= 3 \left({x}^{2} {e}^{x} - 2 \left({e}^{x} x - \setminus \int {e}^{x} \mathrm{dx}\right)\right)$

Using the common integral $\setminus \int {e}^{x} \mathrm{dx} = {e}^{x}$
$= 3 \left({x}^{2} {e}^{x} - 2 \left({e}^{x} x - {e}^{x}\right)\right)$

$= {x}^{3} {e}^{x} - 3 \left({x}^{2} {e}^{x} - 2 \left({e}^{x} x - {e}^{x}\right)\right)$

Adding a constant to the solution,

$= {x}^{3} {e}^{x} - 3 \left({x}^{2} {e}^{x} - 2 \left({e}^{x} x - {e}^{x}\right)\right)$ +C