If $\int \frac{x^3}{x+1} dx = q(x) - log |x + 1| + C$ then q(x) is equal to :

$q(x) = \frac{x^3}{3}-\frac{x^2}{2}+x$

$I = \int \frac{x^3}{x+1} dx$

let y = x + 1

⇒  (y - 1) = x

dx = dy

$I = \int \frac{(y-1)^3}{y}dy = \int \frac{y^3-3y^2+3y-1}{y} dy$

$= \int y^2 - 3y + 3 - \frac{1}{y}  dy$

$=\frac{y^3}{3}-\frac{3y^2}{2}+3y - \log y + C$

$\underbrace{\frac{(x+1)^3}{3}-\frac{3(x+1)^2}{2}+3(x+1)}_{9(x)} - \log (x+1) + C$

$9(x) = (x + 1) \left[\frac{(x+1)^2}{3} - \frac{3(x+1)}{2} + 3\right]$

$=\frac{(x+1)}{6} \left[\frac{2(x^2+2x+1)-9(x+1)+18}{1} \right]$

$=\frac{(x+1)}{6} (2x^2 + 4x + 2 - 9x - 9 + 18)$

$=\frac{(x+1)(2x^2-5x+11)}{6}$

$=\frac{2x^3-5x^2+11x+2x^2-5x+11}{6}$

$=\frac{2x^3-3x^2+6x+11}{6}$

$\underbrace{\frac{x^3}{3}-\frac{x^2}{2}+x}_{9(x)} - \frac{11}{6}$ → considering it a part of constant term