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

`\int (3e^x)/(e^(2x)+2e^x+1)dx=-(3)/(e^x+1)+C`

Explanation:

To find this integral
`\int (3e^x)/(e^(2x)+2e^x+1)dx` you must:

1. Take the constant out:

`3\cdot \int (e^x)/(e^(2x)+2e^x+1)dx`

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

`3\cdot \int (e^x)/(\left(e^x+1\right)^2)dx`

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

`3\cdot \int (e^x)/(\left(u\right)^2)dx`

`u=e^x+1\\du=e^xdx\\dx=(du)/(e^x)`

`3\cdot \int (e^x)/(\left(u\right)^2)(du)/(e^x) \\\\3\cdot \int (1)/(u^2)du`

`3\cdot \int \:u^(-2)du`

4. Apply the Power Rule
`\int x^adx=(x^(a+1))/(a+1),\:\quad \:a\\e -1`

`3\cdot (u^(-2+1))/(-2+1)`

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

`3\cdot (\left(e^x+1\right)^(-2+1))/(-2+1)=3\cdot -\left(e^x+1\right)^(-1)=-(3)/(e^x+1)`

6. Add a constant to the solution

`-(3)/(e^x+1)+C`