Q. 63 Solve each of the integrals in E... [FREE SOLUTION]

Chapter 5: Q. 63 (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.)
$鈭� \frac{x}{\sqrt{3 x + 1}} d x$

Short Answer

Expert verified

The solution of the given integral is $鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{2}{27} {(3 x + 1)}^{3 / 2} - \frac{2}{9} {(3 x + 1)}^{1 / 2} + C$ .

Step by step solution

Step 1. Given Information

Solving the given integrals.

$鈭� \frac{x}{\sqrt{3 x + 1}} d x$

Step 2. Using the substitution method.

Let

$u = 3 x + 1 \frac{d u}{d x} = 3 d u = 3 d x \frac{1}{3} d u = d x$

In this question in which integration by substitution works even though the integrand is not at all in the form $f' (u (x)) u' (x)$ . A clever change of variables will allow us to rewrite the integral so that it can be algebraically simplified.

$3 x = u - 1 x = \frac{u - 1}{3}$

Step 3. This substitution changes the integral into

$鈭� \frac{x}{\sqrt{3 x + 1}} d x = 鈭� \frac{1}{\sqrt{3 x + 1}} 路 x d x 鈭� \frac{x}{\sqrt{3 x + 1}} d x = \frac{1}{3} 鈭� \frac{1}{\sqrt{u}} 路 \frac{u - 1}{3} du 鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{1}{3} 路 \frac{1}{3} 鈭� \frac{1}{u^{1 / 2}} (u - 1) du 鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{1}{9} 鈭� u^{- 1 / 2} (u - 1) du 鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{1}{9} 鈭� (u^{- 1 / 2} 路 u - 1 路 u^{- 1 / 2}) du 鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{1}{9} 鈭� (u^{- 1 / 2 + 1} - u^{- 1 / 2}) du 鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{1}{9} 鈭� (u^{1 / 2} - u^{- 1 / 2}) du$

Step 4. After simplifying.

$鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{1}{9} 鈭� u^{1 / 2} du - \frac{1}{9} 鈭� u^{- 1 / 2} du 鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{1}{9} [\frac{u^{1 / 2 + 1}}{1 / 2 + 1}] - \frac{1}{9} [\frac{u^{- 1 / 2 + 1}}{- 1 / 2 + 1}] + C 鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{1}{9} [\frac{u^{3 / 2}}{3 / 2}] - \frac{1}{9} [\frac{u^{1 / 2}}{1 / 2}] + C 鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{2}{27} u^{3 / 2} - \frac{2}{9} u^{1 / 2} + C 鈭� \frac{x}{\sqrt{3 x + 1}} dx = \frac{2}{27} {(3 x + 1)}^{3 / 2} - \frac{2}{9} {(3 x + 1)}^{1 / 2} + C$

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Solve each of the integrals in Exercises 21鈥�70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)
$鈭� \frac{x}{\sqrt{3 x + 1}} d x$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Using the substitution method.

Step 3. This substitution changes the integral into

Step 4. After simplifying.

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91影视

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�x3x+1dx

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Using the substitution method.

Step 3. This substitution changes the integral into

Step 4. After simplifying.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Geometry

Applied Mathematics

Pure Maths

Logic and Functions

Statistics

Study anywhere. Anytime. Across all devices.

Solve each of the integrals in Exercises 21鈥�70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)
$鈭� \frac{x}{\sqrt{3 x + 1}} d x$