University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Practice Exercises - Page 474: 25


$(\dfrac{1}{3}) \ln |\dfrac{\sqrt {x+1}-1}{\sqrt {x+1}+1}|+c$

Work Step by Step

Consider the integral $\int \dfrac{dx}{x(3 \sqrt {x+1})}$ Let us take the help of the substitution method. $a= \sqrt {x+1} \implies da=\dfrac{dx}{2 \sqrt {x+1}}; dx=(2a) da$ The given integral can be re-written as: $\int \dfrac{dx}{x(3 \sqrt {x+1})}= \dfrac{2}{3}\int \dfrac{da}{(a^2-1)}=\dfrac{1}{3}\int \dfrac{da}{(a-1)}-\dfrac{1}{3}\int \dfrac{da}{(a+1)}$ This implies that $\int [(2x-3)+ \dfrac{x}{x^2+2x-8}]dx=\int (2x-3) dx+(\dfrac{1}{3})\int \dfrac{dx}{x-2}+(\dfrac{2}{3}) \int \dfrac{dx}{x+4}$ We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$ or, $\dfrac{1}{3} \ln |a-1|-\dfrac{1}{3} \ln |a+1|+C=(\dfrac{1}{3}) \ln |\dfrac{\sqrt {x+1}-1}{\sqrt {x+1}+1}|+c$
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