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Chapter 5: Q. 38 (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.)
$鈭� 2 x e^{3 x^{2}} d x$

Short Answer

Expert verified

The solution of the given integral is $鈭� 2 x e^{3 x^{2}} d x = = \frac{1}{3} e^{3 x^{2}} + C$ .

Step by step solution

01

Step 1. Given Information

Solving the given integrals.

$鈭� 2 x e^{3 x^{2}} d x$

02

Step 2. Solving the given integral using substitution method.

Let

$u = 3 x^{2} \frac{d u}{d x} = 6 x d u = 6 x d x \frac{1}{6} d u = x d x$

03

Step 3. This substitution changes the integral into

$鈭� 2 x e^{3 x^{2}} d x = \frac{2}{6} 鈭� e^{u} du 鈭� 2 {xe}^{3 x^{2}} dx = \frac{1}{3} e^{u} + C 鈭� 2 {xe}^{3 x^{2}} dx = \frac{1}{3} e^{3 x^{2}} + C$

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Most popular questions from this chapter

Explain why, if $x = a s e c u$ , then $\sqrt{a^{2} s e c^{2} u 鈭� a^{2}}$ is $鈭� a \tan u$ if $x < 鈭� a$ and is $a \tan u$ if $x > a$ . Your explanation should include a discussion of domains and absolute values.

Solve the integral: $鈭� x \ln (x^{2}) d x$

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.

$鈭� \frac{1}{\sqrt{3 - x^{2}}} d x$

Why is it okay to use a triangle without thinking about the unit circle when simplifying expressions that result from a trigonometric substitution with $x = a \sin u$ or $x = a \tan u$ ? Why do we need to think about the unit circle after trigonometric substitution with $x = a s e c u$ ?

Solve the integral

$鈭� \sqrt{3 - x} \sqrt{x - 1} d x$

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