# integration using partial fractions

$\int \frac{1}{x^{1/2} - x^{1/3}} \, dx = \int \frac{1}{u^3 - u^2} \, (6 u^5 \, du) = 6 \int \frac{u^3}{u - 1} \, du = 6 \int u^2 + u + 1 + \frac{1}{u-1} \, du$