 korporasidn

2021-11-21

Evaluate the integral.
$\int \frac{{\left(1+{e}^{x}\right)}^{2}}{{e}^{x}}dx$ Tionant

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

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