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Chapter 5: Q. 45 (page 417)

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

Short Answer

Expert verified

The solution of the given integral is $鈭� \frac{\ln \sqrt{x}}{x} d x = \frac{{[\ln \sqrt{x}]}^{2}}{4} + C$ .

Step by step solution

01

Step 1. Given Information

Solving the given integrals.

$鈭� \frac{\ln \sqrt{x}}{x} d x$

02

Step 2. Solving the given integral using substitution method.

Let

$u = \ln \sqrt{x} \frac{d u}{d x} = \frac{1}{\sqrt{x}} 路 \frac{1}{2 \sqrt{x}} \frac{d u}{d x} = \frac{1}{2 x} \frac{1}{2} d u = \frac{1}{x} d x$

03

Step 3. This substitution changes the integral into

$鈭� \frac{\ln \sqrt{x}}{x} d x = \frac{1}{2} 鈭� u d u 鈭� \frac{\ln \sqrt{x}}{x} d x = \frac{1}{2} [\frac{u^{1 + 1}}{1 + 1}] + C 鈭� \frac{\ln \sqrt{x}}{x} d x = \frac{1}{2} [\frac{u^{2}}{2}] + C 鈭� \frac{\ln \sqrt{x}}{x} d x = \frac{u^{2}}{4} + C 鈭� \frac{\ln \sqrt{x}}{x} d x = \frac{{[\ln \sqrt{x}]}^{2}}{4} + C$

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Most popular questions from this chapter

Solve given integrals by using polynomial long division to rewrite the integrand. This is one way that you can sometimes avoid using trigonometric substitution; moreover, sometimes it works when trigonometric substitution does not apply.

$鈭� \frac{x^{3} - 3 x^{2} + 2 x - 3}{x^{2} + 1}$ dx

For each integral in Exercises 5鈥�8, write down three integrals that will have that form after a substitution of variables.

$鈭� \frac{1}{\sqrt{u}} d u$

Solve each of the integrals in Exercises 39鈥�74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$鈭� \frac{\sqrt{4 - x^{2}}}{x^{2}} d x$

Explain why it makes sense to try the trigonometric substitution $x = \sec 鈦� u$ if an integrand involves the expression $\sqrt{x^{2} 鈭� 1}$

Solve the integral : $鈭� \frac{x}{e^{x^{2} + 1}} d x$

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