Question

Integrate the following functions:
$x^2\cdot e^{3x}$

Answer 1

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

By applying integration by parts

According to it let $u$ and $v$ be two functions defined in $x$ then,

$\int u\left(x\right)v\left(x\right)dx=u\left(x\right)\int v\left(x\right)dx-\int (\frac{d\left(u\left(x\right)\right)}{dx}\int v\left(x\right)dx)dx$ (1)

Let $v=x^2$ and $u=e^{3x}$

Then according to the formulae in equation (1)

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

Since $\int e^{3x}dx=\frac{e^{3x}}{3}$ and $\frac{d}{dx}\left(x^2\right)=2x$

Equation (2) is reduced to,

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

$=\frac{x^2{e}^{3x}}{3}-\frac{2}{3}{I}_1$ (3)

Solve $I_1$

$I_1=\int x{e}^{3x}dx$

Let $v=x$ and $u=e^{3x}$

Using formula in equation (1)

$\int x{e}^{3x}=x\int e^{x}dx-\int (\frac{d\left(x\right)}{dx}\int e^{3x}dx)dx$ (4)

Since $\int e^{3x}dx=\frac{e^{3x}}{3}$ and $\frac{d\left(x\right)}{dx}=1$

Equation (4 )is reduced to,

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

$=\frac{x^2{e}^{3x}}{3}-\frac{e^{3x}}{9}$

Substitute the value of $I_2$ in equation (3)

$I=\frac{x^2{e}^{3x}}{3}-\frac{2}{3}(\frac{x^2{e}^{3x}}{3}-\frac{e^{3x}}{9})$

$=\frac{x^2{e}^{3x}}{3}-\frac{2x^2{e}^{3x}}{9}+\frac{2e^{3x}}{27}$

$=\frac{x^2{e}^{3x}}{9}+\frac{2e^{3x}}{27}$

Thus, $I=\int x^2{e}^{3x}dx$ is $\frac{x^2{e}^{3x}}{9}+\frac{2e^{3x}}{27}+C$ where $C$ is integration constant.