Calculus of a Single Variable, Chapter 8, 8.1, Section 8.1, Problem 23

intx^2/(x-1)dx
Rewrite the integral as ,
intx^2/(x-1)dx=int(x^2-1+1)/(x-1)dx
=int((x^2-1)/(x-1)+1/(x-1))dx
=int(((x+1)(x-1))/(x-1)+1/(x-1))dx
=int(x+1+1/(x-1))dx
apply the sum rule,
=intxdx+int1dx+int1/(x-1)dx
Apply the power rule and standard integral intdx/x=ln|x|
=(x^(1+1)/(1+1))+x+int1/(x-1)dx
Apply integral substitution u=(x-1) for int1/(x-1)dx
du=dx
int1/(x-1)dx=int(du)/u
=ln(u)
substitute backu=(x-1),
=ln|x-1|
So the final integration and adding a constant C to the solution yields,
=x^2/2+x+ln|x-1|+C