Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - Review - Exercises - Page 577: 5


$$\ln (2t+1)- \ln (t+1)+c $$

Work Step by Step

Given $$\int \frac{dt}{2t^2+3t+1}$$ Since \begin{align*} \frac{1}{2t^2+3t+1}&=\frac{A}{2t +1}+\frac{B}{ t +1}\\ &=\frac{ A(t+1)+ B(2t+1)}{2t^2+3t+1}\\ 1&=A(t+1)+ B(2t+1) \end{align*} At $t=-1 \ \ \Rightarrow B=-1$, and At $t=-1/2 \ \ \Rightarrow A=2$, then $$\frac{1}{2t^2+3t+1} =\frac{2}{2t +1}-\frac{1}{ t +1}$$ Hence \begin{align*} \int\frac{dt}{2t^2+3t+1}&=\int\frac{2dt}{2t +1}-\int\frac{dt}{ t +1}\\ &=\ln (2t+1)- \ln (t+1)+c \end{align*}
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