$∫\frac{x^3}{x+1}dx$ is equal to

$x-\frac{x^2}{2}+\frac{x^3}{3}-log |1+x|+c$

The correct answer is option (1) → $x-\frac{x^2}{2}+\frac{x^3}{3}-log |1+x|+c$

$I=∫\frac{x^3}{x+1}dx$

so dividing numerator by denominator

$∫\frac{x^3+1-1}{x+1}dx$

$⇒∫\frac{x^3+1}{x+1}dx-∫\frac{1}{x+1}dx$

$x^3+1=(x+1)(x^2-x+1)$

$⇒∫\frac{(x+1)(x^2-x+1)}{x+1}dx-∫\frac{1}{x+1}dx$

$⇒∫x^2dx-∫xdx+∫1dx-∫\frac{1}{x+1}dx$

$⇒\frac{x^3}{3}-\frac{x^2}{2}+x+\log |1+x|+c$