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Chapter 5: Q. 41 (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.)
$鈭� x (2^{x^{2}} + 1) d x$

Short Answer

Expert verified

The solution of the given integral is $鈭� x (2^{x^{2} + 1}) d x = \frac{2^{x^{2} + 1}}{2 \log 2} + C$ .

Step by step solution

01

Step 1. Given Information

Solving the given integrals.

$鈭� x (2^{x^{2}} + 1) d x$

02

Step 2. Solving the given integral using substitution method.

Let

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

03

Step 3. This substitution changes the integral into

$鈭� x (2^{x^{2} + 1}) d x = \frac{1}{2} 鈭� 2^{u} du 鈭� x (2^{x^{2} + 1}) d x = \frac{1}{2} 鈭� 2^{u} du 鈭� x (2^{x^{2} + 1}) d x = \frac{1}{2} [\frac{2^{u}}{\log u}] + C 鈭� x (2^{x^{2} + 1}) d x = \frac{2^{u}}{2 \log u} + C 鈭� x (2^{x^{2} + 1}) d x = \frac{2^{x^{2} + 1}}{2 \log 2} + C$

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Most popular questions from this chapter

Solve given definite integral.

${鈭�}_{1 / 4}^{1 / 2} 鈥� \frac{1}{x^{2} \sqrt{1 鈭� x^{2}}} d x$

Why don鈥檛 we ever have cause to use the trigonometric substitution $x = \cos u$ ?

Solve $鈭� \frac{1}{x^{2} + 9} d x$ the following two ways:

(a) with the trigonometric substitution x = 3 tan u;

(b) with algebra and the derivative of the arctangent.

Consider the integral $鈭� \frac{1}{x^{2} \sqrt{1 鈭� x^{2}}} d x$ from the reading at the beginning of the section.

(a) Use the inverse trigonometric substitution $u = \sin^{鈭� 1} 鈦� x$ to solve this integral.

(b) Use the trigonometric substitution $x = \sin 鈦� u$ to solve the integral.

(c) Compare and contrast the two methods used in parts (a) and (b).

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

$鈭� u^{2} d u$

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