Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises: 15

Answer

$\frac{1}{2}(x+1)^{2} - 5(x+1) + 6\ln|(x+1)|$ + C

Work Step by Step

$\int\frac{x^{2}-3x+2}{x+1} dx $ Let $u =x+1$ $\frac{du}{dx}$ = 1 $\frac{du}{1}$ = $dx$ Substitute $u$ and $dx$ into the original equation $\int\frac{x^{2}-3x+2}{u}\frac{du}{1}$ Since $u =x+1$, then $x =(u-1)$ Substitute $x =(u-1)$ into the equation $\int\frac{(u-1)^{2}-3(u-1)+2}{u}\frac{du}{1}$ = $\int\frac{u^{2}+1-2u-3u+3+2}{u}du$ = $\int\frac{u^{2}-5u+6}{u}du$ = $\int\frac{u^{2}}{u}-\frac{5u}{u}+\frac{6}{u}du$ = $\int u-5+\frac{6}{u}du$ = $\int u du - \int 5du + \int\frac{6}{u}du$ = $\frac{u^{2}}{2} - 5u + 6\int\frac{1}{u}du$ = $\frac{u^{2}}{2} - 5u + 6\ln|u|$ + C Since $u =(x+1)$, substituting it back will give you $\frac{1}{2}(x+1)^{2} - 5(x+1) + 6\ln|(x+1)|$ + C
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