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Q. 78

Expert-verified

Calculus

Found in: Page 465

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

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.
$\int \frac{x^{3} - 3 x^{2} + 2 x - 3}{x^{2} + 1}$ dx

$\frac{x^{2}}{2} - 3 x + \frac{1}{2} \ln (|x^{2} + 1|) + C .$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1. Given Information

The integral is as follows.

$\int \frac{x^{3} - 3 x^{2} + 2 x - 3}{x^{2} + 1} d x$

The objective is ton solve the integral.

Step2. Long division

The polynomial long division method is calculated below.

$x - 3 \begin{matrix} x^{2} + 1 x^{3} - 3 x^{2} + 2 x - 3 \\ \frac{x^{3} + x}{- 3 x^{2} + x} \\ \frac{- 3 x^{2} - 3}{x} \end{matrix}$

The expression is in the form of $\frac{x^{3} - 3 x^{2} + 2 x - 3}{x^{2} + 1} = x - 3 + \frac{x}{x^{2} + 1}$

Step3. Solution

The integral is solved below.

$\begin{array}{l} \int \frac{x^{3} - 3 x^{2} + 2 x - 3}{x^{2} + 1} d x \\ = \int (x - 3 + \frac{x}{x^{2} + 1}) d x \\ = \int x d x - \int 3 d x + \int \frac{x}{x^{2} + 1} d x \\ = \frac{x^{2}}{2} - 3 x + \frac{1}{2} \int \frac{1}{u} d u [u = x^{2} + 1, d u = 2 x d x] \\ = \frac{x^{2}}{2} - 3 x + \frac{1}{2} \ln (| u |) + C \frac{x^{2}}{2} - 3 x + \frac{1}{2} \ln (|x^{2} + 1|) + C \end{array}$

Therefore, the value is $\frac{x^{2}}{2} - 3 x + \frac{1}{2} \ln (|x^{2} + 1|) + C .$

Most popular questions for Math Textbooks

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

$\int \frac{x^{2} + 9}{x^{\frac{3}{2}}} d x$

Consider the integral $\int x {(x^{2} - 1)}^{2} d x$ .

(a) Solve this integral by using u-substitution.

(b) Solve the integral another way, using algebra to multiply out the integrand first.

(c) How must your two answers be related? Use algebra to prove this relationship.

Domains and ranges of inverse trigonometric functions: For each function that follows, (a) list the domain and range, (b) sketch a labeled graph, and (c) discuss the domains and ranges in the context of the unit circle.

$f (x) = \sec^{- 1} x$

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

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

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

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

Explain why using trigonometric substitution with $x = a \tan u$ often involves a triangle with side lengths a and x and hypotenuse of length $\sqrt{x^{2} + a^{2}}$

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