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Chapter 7: Problem 51

Use the approaches discussed in this section to evaluate the following integrals. $$\int \frac{e^{x}}{e^{2 x}+2 e^{x}+1} d x$$

Short Answer

Expert verified
Answer: $$-\frac{1}{e^x+1} + C$$

Step by step solution

01

Perform the substitution \(u = e^x\)

Start by letting \(u = e^x\). With this substitution, we need to find the expression for \(dx\) in terms of \(du\). To do this, differentiate both sides with respect to x: $$\frac{d u}{d x}=e^{x}$$ Now, solve for \(dx\): $$d x=\frac{d u}{e^{x}}$$
02

Replace \(e^x\) and \(dx\) in the integral

Using the substitution, rewrite the integral in terms of u: $$\int \frac{e^{x}}{e^{2 x}+2 e^{x}+1} d x = \int \frac{1}{u^2 + 2u + 1} \, d u$$
03

Complete the square in the denominator

The denominator can be simplified by completing the square. Let's take a look at the quadratic in the denominator: $$u^2 + 2u + 1$$ This quadratic can be written as the square of a binomial: $$(u+1)^2$$ Now, we can rewrite the integral as: $$\int \frac{1}{(u+1)^2} \, d u$$
04

Perform the integration

The integral is now a basic power function. To solve it, we can use the power rule of integration: $$\int x^n \, dx = \frac{x^{n+1}}{n+1}$$ Apply the power rule to our integral: $$\int \frac{1}{(u+1)^2} \, d u = -\frac{1}{u+1} + C$$ where C is the constant of integration.
05

Substitute back to x

Finally, replace u with the original expression, \(u=e^x\): $$-\frac{1}{u+1} + C = -\frac{1}{e^x+1} + C$$ The evaluated integral is: $$\int \frac{e^{x}}{e^{2 x}+2 e^{x}+1} d x = -\frac{1}{e^x+1} + C$$

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