Question

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

Short answer

The required answer is $\left(-\frac{x}{e^x}\right)-\frac{x}{2e^x}-\frac{1}{2}{e}^{-x}+c$.

Step-by-step solution

Step 1. Given information. 

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

Step 2. Solve the integration by parts . 

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

$$\begin{gathered} u=x^2 \\ du=2xdx \\ \frac{du}{2}=xdx \end{gathered}$$

and

$$\begin{gathered} dv=\frac{dx}{e^x} \\ v=\int \frac{dx}{e^x} \\ v=-\frac{1}{e^x} \end{gathered}$$

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

role="math" localid="1648713324601" $$\begin{gathered} \int \frac{x^2}{e^x}dx \\ =\left[x^2\left(-\frac{1}{e^x}\right)-\frac{1}{2}\int x\left(-\frac{1}{e^x}\right)dx\right] \\ =\left(-\frac{x}{e^x}\right)+\frac{1}{2}\int \frac{x}{e^x}dx \\ =\left(-\frac{x}{e^x}\right)+\frac{1}{2}\left[x\left(-\frac{1}{e^x}\right)-\int \left(-\frac{1}{e^x}\right)dx\right] \\ =\left(-\frac{x}{e^x}\right)+\frac{1}{2}\left[\left(-\frac{x}{e^x}\right)+\int e^{-x}dx\right] \end{gathered}$$

Step 3. Integration by parts.  

We have, $$\begin{gathered} u=-x \\ du=-dx \end{gathered}$$

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