∫ xdx / (x+1)³
= ∫ (x+1-1) / (x+1)³ dx
= ∫ [ (x+1) / (x+1)³ - 1/ (x+1)³ ] dx
= ∫ [ 1/(x+1)² - 1/(x+1)³ ] dx
=-1/(x+1)+1/[2(x+1)²] + C
=-(x+½)/(x+1)² + C
∫ [x³ / (x+1)] dx
=∫ { [ (x+1)³ - 3x² - 3x - 1] / (x+1)} dx
=∫ [ (x+1)² - 3x - 1/(x+1) ] dx
=1/3 (x+1)³ - 3/2 x² - ln(x+1) + C