Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 501: 40

Answer

$\displaystyle \ln|x+2\sqrt{x+3}|+\frac{1}{2}\ln\left|\frac{\sqrt{x+3}+1}{2}+1\right|-\frac{1}{2}\ln\left|\frac{\sqrt{x+3}+1}{2}-1\right|+C$

Work Step by Step

$\displaystyle \int\frac{dx}{2\sqrt{x+3}+x}=\quad\left[\begin{array}{ll} t=\sqrt{x+3} & x=t^{2}-3\\ t^{2}=x+3 & dx=2tdt \end{array}\right]$ $=\displaystyle \int\frac{2t}{2t+t^{2}-3}dt=\int\frac{2t+2-2}{t^{2}+2t-3}dt$ $=\displaystyle \int\frac{2t+2}{t^{2}+2t-3}dt-2\int\frac{1}{t^{2}+2t+1-4}dt$ $=\displaystyle \ln|t^{2}+2t-3|-2\int\frac{1}{(t+1)^{2}-2^{2}}dt$ $\color{blue}{ \displaystyle \begin{aligned}\int\frac{dx}{x^{2}-a^{2}}&=\displaystyle \frac{1}{2a}\int\left(\frac{1}{x-a}-\frac{1}{x+a}\right)dx\\&=\displaystyle \frac{1}{2a}(\ln|x-a|-\ln|x+a|)+C\end{aligned}}$ $=\displaystyle \ln|t^{2}+2t-3| +\frac{1}{2}\left(\ln\left|\frac{t+1}{2}+1\right|-\ln\left|\frac{t+1}{2}-1\right|\right)+C$ $=\displaystyle \ln|x+2\sqrt{x+3}|+\frac{1}{2}\ln\left|\frac{\sqrt{x+3}+1}{2}+1\right|-\frac{1}{2}\ln\left|\frac{\sqrt{x+3}+1}{2}-1\right|+C$
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