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
