## Calculus 10th Edition

Published by Brooks Cole

# Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.2 Exercises: 16

#### Answer

$(x-2)^{2} + 15(x-2) + 19\ln|(x-2)|$ + C

#### Work Step by Step

$\int\frac{2x^{2}+7x-3}{x-2} dx$ Let $u =x-2$ $\frac{du}{dx}$ = 1 $\frac{du}{1}$ = $dx$ Substitute $u$ and $dx$ into the original equation $\int\frac{2x^{2}+7x-3}{u}\frac{du}{1}$ Since $u =x-2$, Hence, $x =(u+2)$ Substitute $x =(u+2)$ into the equation $\int\frac{2(u+2)^{2}+7(u+2)-3}{u}\frac{du}{1}$ = $\int\frac{2(u^{2}+4+4u)+7u+14-3}{u}du$ = $\int\frac{2u^{2}+8+8u+7u+14-3}{u}du$ = $\int\frac{2u^{2}+15u+19}{u}du$ = $\int(\frac{2u^{2}}{u}+\frac{15u}{u}+\frac{19}{u})du$ = $\int (2u+15+\frac{19}{u})du$ = $\int 2u du + \int 15du + \int\frac{19}{u}du$ = $\frac{2u^{2}}{2} + 15u + 19\int\frac{1}{u}du$ = $u^{2} + 15u + 19\ln|u|$ + C Since $u =(x-2)$, substituting it back will give you $(x-2)^{2} + 15(x-2) + 19\ln|(x-2)|$ + C

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