Evaluate $\int \frac{x^2}{x-1} \,dx$ Evaluate $\int \frac{x^2}{x-1} \,dx$
(A) $2x^2+x+\ln|x-1|+C$
(B) $\frac{x^2}{2}+x+\ln|x+1|+C$
(C) $\frac{x^2}{2}+x+\ln|x-1|+C$
(D) $x^2+x+\ln|x-1|+C$
My attempt :
Let $u=x-1$, so : $du=dx$ and $u+1$
$\int \frac{(u+1)^2}{u}\,du\\
=\int u +2+\frac{1}{u}\,du\\
=\frac{u^2}{2}+2u+\ln|u|+C\\
=\frac{(x-1)^2}{2}+2(x-1)+\ln|x-1|+C$
Simplify :
$\frac{x^2-3}{2}+x+\ln|x-1|+C$
It's not on the option.

Your answer is right except for the constant I changed.
$$\frac{x^2-3}{2}+x+\ln|x-1|+C_1 = \frac{x^2}{2}+x+\ln|x-1|+C_1-\frac{3}{2} = \frac{x^2}{2}+x+\ln|x-1|+C$$
