Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - Review - Exercises - Page 578: 19

Answer

\[\frac{1}{18}\ln(9x^2+6x+5)+\frac{1}{9}\tan^{-1}\left[\frac{1}{2}(3x+1)\right]+C\]

Work Step by Step

Let \[I=\int\frac{x+1}{9x^2+6x+5}dx\] \[x+1=A\frac{d}{dx}[9x^2+6x+5]+B\] \[x+1=A[18x+6]+B\] Comparing like coefficients \[18A=1\Rightarrow A=\frac{1}{18}\] and \[6A+B=1\Rightarrow B+\frac{1}{3}=1\Rightarrow B=\frac{2}{3}\] \[\Rightarrow I=\frac{1}{18}\int\frac{18x+6}{9x^2+6x+5}dx+\frac{2}{3}\int\frac{dx}{9x^2+6x+5}\;\;\;\;\ldots (1)\] Let \[I_1=\int\frac{18x+6}{9x^2+6x+5 }dx\] Substitute $\;t=9x^2+6x+5\;\;\;\ldots (2)$ \[\Rightarrow dt=(18x+6)dx\] \[I_1=\int\frac{dt}{t}=\ln t+C_1\] Where $C_1$ is constant of integration From (2) \[I_1=\ln(9x^2+6x+5)+C_1\;\;\ldots (3)\] Let \[I_2=\int\frac{dx}{9x^2+6x+5}\] \[I_2=\frac{1}{9}\int\frac{dx}{x^2+\frac{2x}{3}+\frac{5}{9}}\] \[I_2=\frac{1}{9}\int\frac{dx}{x^2+\frac{2x}{3}+\frac{5}{9} +\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2 }\] \[I_2=\frac{1}{9}\int\frac{dx}{\left(x+\frac{1}{3}\right)^2+\left(\frac{2}{3}\right)^2}\] \[I_2=\frac{1}{9}\left[\frac{1}{\left(\frac{2}{3}\right)}\tan^{-1}\left(\frac{x+\frac{1}{3}}{\frac{2}{3}}\right)\right]+C_2\] Where $C_2$ is constant of integration \[I_2=\frac{1}{6}\tan^{-1}\left(\frac{3x+1}{2}\right)+C_2\;\;\;\ldots (4)\] Using (3) and (4) in (1) \[I=\frac{1}{18}\ln(9x^2+6x+5)+\frac{1}{9}\tan^{-1}\left[\frac{1}{2}(3x+1)\right]+C\] Where $C=C_1+C_2$ Hence \[I=\frac{1}{18}\ln(9x^2+6x+5)+\frac{1}{9}\tan^{-1}\left[\frac{1}{2}(3x+1)\right]+C\]
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