University Calculus: Early Transcendentals (3rd Edition)

$x+ \ln |x-1|-\ln |x|+C$
Consider the integral $\int \dfrac{x^3+1}{x^3-x}dx$ The given integral can be re-written as: $\int \dfrac{x^3+1}{x^3-x}dx=\int1+\dfrac{x+1}{x^3-x}dx=\int 1 dx+\int \dfrac{1}{x(x-1)}dx$ This implies that $\int 1 dx+\int \dfrac{1}{x(x-1)}dx=x+ \ln |x-1|-\ln |x|+C$