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Chapter 5: Q. 76 (page 418)

Solve each of the integrals in Exercises 21鈥�70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)
${鈭�}_{2}^{4} \frac{e^{2 x}}{1 + e^{2 x}} d x$

Short Answer

Expert verified

The solution of the given integral is ${鈭�}_{2}^{4} \frac{e^{2 x}}{1 + e^{2 x}} d x = [\ln e^{8} - \ln e^{4}]$ .

Step by step solution

01

Step 1. Given Information

Solving the given integrals.

${鈭�}_{2}^{4} \frac{e^{2 x}}{1 + e^{2 x}} d x$

02

Step 2. Using the substitution method.

Let

$u = 1 + e^{2 x} \frac{d u}{d x} = e^{2 x} d u = e^{2 x} d x$

03

Step 3. Using the information in equations, we can change variables completely:

${鈭�}_{2}^{4} \frac{e^{2 x}}{1 + e^{2 x}} d x = {鈭�}_{x = 2}^{x = 4} \frac{1}{u} d u {鈭�}_{2}^{4} \frac{e^{2 x}}{1 + e^{2 x}} d x = {[\ln u]}_{x = 2}^{x = 4} {鈭�}_{2}^{4} \frac{e^{2 x}}{1 + e^{2 x}} d x = {[\ln e^{2 x}]}_{x = 2}^{x = 4} {鈭�}_{2}^{4} \frac{e^{2 x}}{1 + e^{2 x}} d x = [\ln e^{2 脳 4} - \ln e^{2 脳 2}] {鈭�}_{2}^{4} \frac{e^{2 x}}{1 + e^{2 x}} d x = [\ln e^{8} - \ln e^{4}]$

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Most popular questions from this chapter

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