Evaluate the integral. ∫(1+ex)2exdx

Step 1
The given integral is ${\displaystyle \int \frac{{(1+e^x)}^2}{e^x}dx}$.
Step 2
Evaluate the given integral as shown below:
${\displaystyle \int \frac{{(1+e^x)}^2}{e^x}dx=\int \frac{1+2e^x+e^{2x}}{e^x}dx}$
${\displaystyle =\int \frac{1}{e^x}+2+e^{x}dx}$
${\displaystyle =\int \frac{1}{e^x}dx+\int 2dx+\int e^{x}dx}$
${\displaystyle =-\frac{1}{e^x}+2x+e^x+C}$

Step 1: Expand.
${\displaystyle \int \frac{1+2e^x+e^{2x}}{e^x}dx}$
Step 2: Split fraction.
${\displaystyle \int \frac{1}{e^x}+\frac{2e^x}{e^x}+\frac{e^{2x}}{e^x}dx}$
Step 3: Use Sum Rule: ${\displaystyle \int f\left(x\right)+g\left(x\right)dx=\int f\left(x\right)dx+\int g\left(x\right)dx}$.
${\displaystyle \int \frac{1}{e^x}dx+\int 2dx+\int e^{x}dx}$
Step 4: Use Integration by Substitution on ${\displaystyle \int \frac{1}{e^x}dx}$.
Let ${\displaystyle u=\frac{1}{e^x},du=-\frac{1}{e^x}dx,thendx=-e^{\ln \frac{1}{u}}du}$
Step 5: Using u and du above, rewrite ${\displaystyle \int \frac{1}{e^x}dx}$.
${\displaystyle \int u\times -e^{\ln \frac{1}{u}}du}$
Step 6: Use this rule: ${\displaystyle \int adx=ax+C}$.
-u
Step 7: Substitute ${\displaystyle u=\frac{1}{e^x}}$ back into the original integral.
${\displaystyle -\frac{1}{e^x}}$
Step 8: Rewrite the integral with the completed substitution.
${\displaystyle -\frac{1}{e^x}+\int 2dx+\int e^{x}dx}$
Step 9: Use this rule: ${\displaystyle \int adx=ax+C}$.
${\displaystyle -\frac{1}{e^x}+2x+\int e^{x}dx}$
Step 10: The integral of ${\displaystyle e^{x}is{e}^x}$.
${\displaystyle -\frac{1}{e^x}+2x+e^x}$
Step 11: Add constant.
${\displaystyle -\frac{1}{e^x}+2x+e^x+C}$