Question

Solve the integral:$\int {(x-e^x)}^{2}dx$

Short answer

The required answer is$\frac{x^3}{3}-2x{e}^x+2e^x+\frac{1}{2}{e}^{2x}+c$.

Step-by-step solution

Step 1. Given information. 

We have given integral is$\int {\left(x-e^x\right)}^{2}dx$.

Step 2. Solve the integration by parts . 

We have,

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

Step 3. Integration by parts.

We have,

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

and

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

The formula of integration by parts is $\int udv=uv-\int vdu$
$$\begin{gathered} \frac{x^3}{3}-\left(2\left(x{e}^x-\int e^{x}dx\right)\right)+\int e^{2x}dx \\ =\frac{x^3}{3}-2x{e}^x+2e^x+\int e^{2x}dx \\ =\frac{x^3}{3}-2x{e}^x+2e^x+\left(\frac{1}{2}\int e^{2x}dx\right) \\ =\frac{x^3}{3}-2x{e}^x+2e^x+\frac{1}{2}{e}^{2x}+c \end{gathered}$$