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Chapter 5: Q. 77 (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.)
${鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x$

Short Answer

Expert verified

The solution of the given integral is ${鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x = \frac{蟺}{8}$ .

Step by step solution

01

Step 1. Given Information

Solving the given integrals.

${鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x$

02

Step 2. Using the substitution method.

${鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x = {鈭�}_{0}^{1} \frac{x}{1 + {(x^{2})}^{2}} d x$

Let

role="math" localid="1649086854342" $u = x^{2} \frac{d u}{d x} = 2 x d u = 2 x d x \frac{1}{2} d u = x d x$

03

Step 3. We will now write the limits of integration in terms of the new variable u.

When $x = 0$ , we have

$u = x^{2} u = 0^{2} u = 0$

When role="math" localid="1649086870398" $x = 1$ , we have

$u = x^{2} u = {(1)}^{2} u = 1$

04

Step 4. Using the information in equations, we can change variables completely:

${鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x = \frac{1}{2} {鈭�}_{0}^{1} \frac{1}{1 + u^{2}} d u {鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x = \frac{1}{2} {[\tan^{- 1} x]}_{0}^{1} {鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x = \frac{1}{2} [\tan^{- 1} 1 - \tan^{- 1} 0] {鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x = \frac{1}{2} [\frac{蟺}{4} - 0^{o}] {鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x = \frac{1}{2} 路 \frac{蟺}{4} {鈭�}_{0}^{1} \frac{x}{1 + x^{4}} d x = \frac{蟺}{8}$

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