Answer: Repeated quadratic factors Refer to the summary

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

ISBN: 9780321570567 2

Solution for problem 68E Chapter 7.4

Calculus: Early Transcendentals | 1st Edition

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Problem 68E

Repeated quadratic factors

Refer to the summary box on p. 483 and evaluate the following integrals.

Step-by-Step Solution:

Step 1 of 3

Answer

Given integral is $\int\limits_{\ }^{\ }\frac{dx}{(x+1)({x}^{2}+2x+2{)}^{2}}\$

We have to evaluate the given integral

$\int\limits_{\ }^{\ }\frac{dx}{(x+1)({x}^{2}+2x+2{)}^{2}}\ =\int\limits_{\ }^{\ }\frac{1}{(x+1)({x}^{2}+2x+1{)}^{2}}\ dx$

Put $u={x}^{\ }+1\Rightarrow{}du=\ dx$

$\int\limits_{\ }^{\ }\frac{dx}{(x+1)({x}^{2}+2x+2{)}^{2}}\ =\int\limits_{\ }^{\ }\frac{1}{u({x}^{2}+2x+1{)}^{2}}\ du$

$=\int\limits_{\ }^{\ }\frac{1}{u({x}^{\ }(x+2)+1{)}^{2}}\ du$

$=\int\limits_{\ }^{\ }\frac{1}{u({(u-1)}^{\ }(u-1+2)+1{))}^{2}}\ du$

$=\int\limits_{\ }^{\ }\frac{1}{u({(u-1)}^{\ }(u+2{))}^{2}}\ du$

$\int\limits_{\ }^{\ }\frac{dx}{(x+1)({x}^{2}+2x+2{)}^{2}}$ $=\int\limits_{\ }^{\ }\frac{1}{u({u}^{2}+1{)}^{2}}\ du$ …(1)

To evaluate the integral $\int\limits_{\ }^{\ }\frac{1}{u({u}^{2}+1{)}^{2}}\ du$

we substitute

$v={u}^{2}+1\Rightarrow{}dv=2u\ du\$

$\int\limits_{\ }^{\ }\frac{1}{u({u}^{2}+1{)}^{2}}\ du=\int\limits_{\ }^{\ }\frac{1}{2(v-1){v}^{2}}dv$

$=\frac{1}{2}[-\int\limits_{\ }^{\ }\frac{1}{{v}^{2}}dv-\int\limits_{\ }^{\ }\frac{1}{v}\ dv+\int\limits_{\ }^{\ }\frac{1}{v-1}dv]$ ..(partial fraction )

$\int\limits_{\ }^{\ }\frac{1}{u({u}^{2}+1{)}^{2}}\ du$ $=\frac{1}{2}[-(-\frac{1}{v})-ln\ v+ln\ (v-1)]$

Substitute back $v={u}^{2}+1\$

$\int\limits_{\ }^{\ }\frac{1}{u({u}^{2}+1{)}^{2}}\ du$ $=\frac{1}{2}[(\frac{1}{{u}^{2}+1})-ln\ ({u}^{2}+1)+ln\ ({u}^{2}+1-1)]$

$=\frac{1}{2}[(\frac{1}{{u}^{2}+1})-ln\ ({u}^{2}+1)+ln\ ({u}^{2})]$

$=\frac{1}{2}[(\frac{1}{{u}^{2}+1})-ln\ ({u}^{2}+1)+2ln\ ({u}^{\ })]$

Substitute back $u={\ }^{\ }x+1\$

$\int\limits_{\ }^{\ }\frac{1}{u({u}^{2}+1{)}^{2}}\ du=\frac{1}{2}[(\frac{1}{{(x+1)}^{2}+1})-ln\ \left|{{(x+1)}^{2}+1}\right|+2ln\left|{\ {x+1}^{\ }}\right|]+C$

Therefore

$\int\limits_{\ }^{\ }\frac{1}{u({u}^{2}+1{)}^{2}}\ du=\frac{1}{2}[(\frac{1}{{(x+1)}^{2}+1})-ln\ \left|{{(x+1)}^{2}+1}\right|+2ln\left|{\ {x+1}^{\ }}\right|]+C$

Step 2 of 3

Chapter 7.4, Problem 68E is Solved

Step 3 of 3

Textbook: Calculus: Early Transcendentals

Edition: 1

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

ISBN: 9780321570567

This full solution covers the following key subjects: box, evaluate, factors, Integrals, quadratic. This expansive textbook survival guide covers 112 chapters, and 5248 solutions. The answer to “Repeated quadratic factors Refer to the summary box on p. 483 and evaluate the following integrals.” is broken down into a number of easy to follow steps, and 16 words. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. Since the solution to 68E from 7.4 chapter was answered, more than 242 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. The full step-by-step solution to problem: 68E from chapter: 7.4 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM.