# Find the indefinite integral. Hint: Let v = e - u + 2u. (Remember to use ln(|u|) where...

## Question:

Find the indefinite integral. Hint: Let v = e - u + 2u. (Remember to use ln(|u|) where appropriate.)

{eq}\int \frac{e^{-u} - 2}{e^{-u} + 2u} du {/eq}

## Integration by Substitution

In the method of integration by substitution, the variable is substituted with another function such that the complicated integral becomes easier.

The chain rule in differentiation states that {eq}\displaystyle \frac{d}{dx} f(g(x))=f'(g(x))g'(x) {/eq}

We need to know that {eq}\displaystyle \int \frac{f'(x)}{f(x)} =\ln |f(x)| +C , \int \frac{1}{x} dx =\ln |x|+C {/eq} where {eq}\displaystyle C {/eq} is the constant of integration.

## Answer and Explanation:

Let {eq}\displaystyle I =\int \frac{e^{-u}-2 }{e^{-u}+2u } du {/eq}

Put {eq}\displaystyle e^{-u} +2u =v {/eq}

Differentiating, we get

{eq}\displaystyle \frac{d}{du}(e^{-u} +2u) =\frac{d}{du}v {/eq}

Using the chain rule, we get

{eq}\displaystyle e^{-u} \frac{d}{du} (-u) +2 =\frac{dv}{du} \\ \Rightarrow du ( -e^{-u} +2 ) = dv \\ \Rightarrow du (e^{-u} -2) = -dv {/eq}

Substituting, we get

{eq}\displaystyle I= \int \frac{-dv}{v} \\ \Rightarrow I = - \ln |v| +C {/eq}

As {eq}\displaystyle e^{-u} +2u =v {/eq}, we get

{eq}\displaystyle I = -\ln |e^{-u} +2u|+C {/eq}

Therefore, {eq}\displaystyle \mathbf{\int \frac{e^{-u}-2 }{e^{-u}+2u } du=-\ln |e^{-u} +2u|+C} {/eq}

#### Learn more about this topic: How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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