Question

Evaluate the following integral. Integral from nothing to nothing StartFraction 3 e Superscript x Over e Superscript 2 x Baseline plus 2 e Superscript x Baseline plus 1 EndFraction dx∫ 3ex e2x+2ex+1dx Integral from nothing to nothing StartFraction 3 e Superscript x Over e Superscript 2 x Baseline plus 2 e Superscript x Baseline plus 1 EndFraction dx∫ 3ex e2x+2ex+1dxequals

Answer 1

Answer:

$\int \frac{3e^x}{e^{2x}+2e^x+1}dx=-\frac{3}{e^x+1}+C$

Step-by-step explanation:

To find this integral $\int \frac{3e^x}{e^{2x}+2e^x+1}dx$ you must:

1. Take the constant out:

$3\cdot \int \frac{e^x}{e^{2x}+2e^x+1}dx$

2. Factor ${e^{2x}+2e^x+1}=(e^{x}+1)^2$

$3\cdot \int \frac{e^x}{\left(e^x+1\right)^2}dx$

3. Apply u-substitution $u=e^x+1$

$3\cdot \int \frac{e^x}{\left(u\right)^2}dx$

$u=e^x+1\\du=e^xdx\\dx=\frac{du}{e^x}$

$3\cdot \int \frac{e^x}{\left(u\right)^2}\frac{du}{e^x} \\\\3\cdot \int \frac{1}{u^2}du$

$3\cdot \int \:u^{-2}du$

4. Apply the Power Rule $\int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$3\cdot \frac{u^{-2+1}}{-2+1}$

5. Substitute back $u=e^x+1$

$3\cdot \frac{\left(e^x+1\right)^{-2+1}}{-2+1}=3\cdot -\left(e^x+1\right)^{-1}=-\frac{3}{e^x+1}$

6. Add a constant to the solution

$-\frac{3}{e^x+1}+C$