If ∫(x^3)/(x+1) dx = q(x) - log|x + 1| + C then q(x) is equal to:

The correct answer is Option (4) → \(q(x) = \displaystyle\frac{x^3}{3} - \frac{x^2}{2} + x\)

The numerator has a higher degree than the denominator, so divide first. Adding and subtracting 1 and using \(x^3 + 1 = (x + 1)(x^2 - x + 1)\):

    \( \displaystyle \frac{x^3}{x+1} = \frac{(x^3 + 1) - 1}{x+1} \)

    \( \displaystyle = x^2 - x + 1 - \frac{1}{x+1} \)

Integrating term by term:

    \( \displaystyle \int\frac{x^3}{x+1}\,dx = \frac{x^3}{3} - \frac{x^2}{2} + x - \log|x + 1| + C \)

Comparing with \(q(x) - \log|x + 1| + C\):

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