Question

Solve the integral:$\int x^2{e}^{3x}dx$.

Short answer

The required answer is$\frac{x^2{e}^{3x}}{3}-\frac{2}{9}x{e}^{3x}+\frac{2}{27}{e}^{3x}+c$.

Step-by-step solution

Step 1. Given information. 

We have given integral is$\int x^2{e}^{3x}dx$.

Step 2. Solve the integration by parts . 

We have, $u=x^2,dv=e^{3x}dx$

localid="1648652451162">u=x2du=2xdx

and

$$\begin{gathered} dv=e^{3x}dx \\ v=\int e^{3x}dx \\ v=\frac{e^{3x}}{3} \end{gathered}$$

The formula of integration by parts is $\int udv=uv-\int vdu$

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

Step 3. Integration by parts. 

We have, $u=x,dv=e^{3x}dx$

$$\begin{gathered} u=x \\ du=dx \end{gathered}$$

and

$$\begin{gathered} dv=e^{3x}dx \\ v=\int e^{3x}dx \\ v=\frac{e^{3x}}{3} \end{gathered}$$

So,

$$\begin{gathered} \frac{1}{3}{x}^2{e}^{3x}-\frac{2}{3}\int x\frac{e^{3x}}{3}dx \\ =\frac{1}{3}{x}^2{e}^{3x}-\frac{2}{3}\left[\frac{1}{3}x{e}^{3x}-\frac{1}{3}\int e^{3x}dx\right] \\ =\frac{1}{3}{x}^2{e}^{3x}-\frac{2}{9}x{e}^{3x}+\frac{2}{27}{e}^{3x}+c \end{gathered}$$