Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.4 Integration of Rational Functions by Partial Fractions - 7.4 Exercises - Page 541: 50


$$\int \frac{e^{x}}{\left(e^{x}-2\right)\left(e^{2 x}+1\right)} d x=\frac{1}{5} \ln \left|e^{x}-2\right|-\frac{1}{10} \ln \left(e^{2 x}+1\right)-\frac{2}{5} \arctan e^{x}+C$$

Work Step by Step

Given $$\int \frac{e^{x}}{\left(e^{x}-2\right)\left(e^{2 x}+1\right)} d x$$ Let \begin{aligned} u &=e^{x} \\ d u &=e^{x} d x \\ &\text {so we get}\\ I& = \int \frac{e^{x}}{\left(e^{x}-2\right)\left(e^{2 x}+1\right)} d x \\ &=\int \frac{1}{(u-2)\left(u^{2}+1\right)} d u \end{aligned} Since \begin{array}{l}{\frac{1}{(u-2)\left(u^{2}+1\right)}=\frac{A}{u-2}+\frac{B u+C}{u^{2}+1}} \\ {\quad 1=A\left(u^{2}+1\right)+(B u+C)(u-2)} \\ {\text { at } u=2 : \quad 1=A(4+1)+0} \\ {\quad A=\frac{1}{5}} \\ {\quad \text { at } u=0 : \quad 1=\frac{1}{5}(0+1)+(0+C)(0-2)}\\ {}{\quad \quad \quad \quad \quad \quad \quad 1=\frac{1}{5}-2 C} \\ {\quad \quad \quad \quad \quad \quad \quad\frac{4}{5}=-2 C} \\ {\quad C=-\frac{2}{5}} \\ {\quad \text { at } u=1 : \quad 1=2 A-(B+C)}\\ {}{\quad \quad \quad \quad \quad \quad \quad 1=\frac{2}{5}+\frac{2}{5}-B} \\ {\quad \quad \quad \quad \quad \quad \quad-\frac{1}{5}=-B} \\ {\quad B=-\frac{1}{5}} \end{array} So, we get \begin{aligned}I&= \int \frac{1}{(u-2)\left(u^{2}+1\right)} d u \\&=\int\left(\frac{1 / 5}{u-2}+\frac{(-1 / 5) u-(2 / 5)}{u^{2}+1}\right) d u \\ &=\int\left(\frac{1 / 5}{u-2}+\frac{-u-2}{5\left(u^{2}+1\right)}\right) d u \\ &=\int\left(\frac{1 / 5}{u-2}-\frac{u}{5\left(u^{2}+1\right)}-\frac{2}{5\left(u^{2}+1\right)}\right) d u \\ &=\frac{1}{5} \ln |u-2|-\frac{1}{5} \cdot \frac{1}{2} \ln \left|u^{2}+1\right|-\frac{2}{5} \arctan u+C \\ &=\frac{1}{5} \ln \left|e^{x}-2\right|-\frac{1}{10} \ln \left(e^{2 x}+1\right)-\frac{2}{5} \arctan e^{x}+C \end{aligned}
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