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
