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Chapter 5: Q. 34 (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.)
$鈭� \frac{\sqrt{x} + 1}{x} d x$

Short Answer

Expert verified

The solution of the given integral is $鈭� \frac{\sqrt{x} + 1}{x} d x = 2 \sqrt{x} + \ln x + C$ .

Step by step solution

01

Step 1. Given Information

Solving the given integrals.

$鈭� \frac{\sqrt{x} + 1}{x} d x$

02

Step 2. Solving the given integral using algebra.

$鈭� \frac{\sqrt{x} + 1}{x} d x = 鈭� (\frac{\sqrt{x}}{x} + \frac{1}{x}) d x 鈭� \frac{\sqrt{x} + 1}{x} d x = 鈭� (\frac{1}{\sqrt{x}} + \frac{1}{x}) d x 鈭� \frac{\sqrt{x} + 1}{x} d x = 鈭� (\frac{1}{x^{1 / 2}} + \frac{1}{x}) d x 鈭� \frac{\sqrt{x} + 1}{x} d x = 鈭� (x^{- 1 / 2} + \frac{1}{x}) d x$

03

Step 3. After simplifying

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

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Most popular questions from this chapter

Solve given definite integral.

${鈭�}_{3}^{5} 鈥� \sqrt{x^{2} 鈭� 9} d x$

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For each function u(x) in Exercises 9鈥�12, write the differential du in terms of the differential dx.

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