Answer
$\int\frac{x^3-x^2}{3x^4-4x^3}dx=\frac{1}{12}\ln|3x^4-4x^3|+C$
Work Step by Step
Substitution:
$u=3x^4-4x^3$
$\frac{du}{dx}=12x^3-12x^2=12(x^3-x^2)$
$dx=\frac{1}{12(x^3-x^2)}du$
$\int\frac{x^3-x^2}{3x^4-4x^3}dx=\int\frac{x^3-x^2}{u}\frac{1}{12(x^3-x^2)}du=\int\frac{1}{12}\frac{1}{u}du=\frac{1}{12}\ln|u|+C=\frac{1}{12}\ln|3x^4-4x^3|+C$
