Q.$\int \dfrac{x^3}{x+1}\,dx$ is equal to
(A) $x + \dfrac{x^2}{2} + \dfrac{x^3}{3} - \log|1-x| + C$
(B) $x + \dfrac{x^2}{2} - \dfrac{x^3}{3} - \log|1-x| + C$
(C) $x - \dfrac{x^2}{2} - \dfrac{x^3}{3} - \log|1+x| + C$
(D) $x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \log|1+x| + C$

Concept: Polynomial long division before integration.

Since the numerator’s degree (3) is higher than the denominator’s (1), divide first:

Now integrate term by term:

So the integral is: …

The key insight: when the numerator's degree is higher than the denominator's, you must divide first. A rational function like $\frac{x^3}{x+1}$ is not in a directly integrable form — but after division, it becomes a simple polynomial plus a proper fraction, each piece easy to integrate.

Let’s walk through it.

1. Check degrees and decide the method

   The numerator $x^3$ is degree 3, denominator $x+1$ is degree 1. Since $3 \ge 1$, we perform polynomial long division (or synthetic division) to rewrite the fraction.

2. Perform the division

   Divide $x^3$ by $x+1$:

   • $x^3 \div x = x^2$, multiply back: $x^2(x+1) = x^3 + x^2$, subtract from $x^3$ (which has no $x^2$ term, so treat as $x^3 + 0x^2$): $(x^3) - (x^3 + x^2) = -x^2$.
   • Bring down the next term (0x): $-x^2 \div x = -x$, multiply: $-x(x+1) = -x^2 - x$, subtract: $(-x^2) - (-x^2 - x) = x$.
   • Bring down the constant (0): $x \div x = 1$, multiply: $1(x+1) = x+1$, subtract: $x - (x+1) = -1$.

   So the quotient is $x^2 - x + 1$ and the remainder is $-1$. Therefore:

3. Integrate term by term Now the integral becomes:

Each piece is standard: …

Method: Polynomial long division for an improper fraction

Whenever the numerator's degree is $\ge$ the denominator's, divide before integrating.

Steps

Step 1: Check the degrees.

If $\deg(\text{num})\ge\deg(\text{den})$, do long division to get quotient $+\dfrac{\text{remainder}}{\text{den}}$.

Step 2: Divide.

For $\dfrac{x^3}{x+1}$, division gives $x^2-x+1-\dfrac{1}{x+1}$. …

Common Mistakes

Mistake 1: Integrating $\frac{x^3}{x+1}$ without dividing.

Why it's wrong: it is an improper fraction ($\deg 3>\deg 1$) with no direct antiderivative. Correct approach: divide first to get $x^2-x+1-\frac{1}{x+1}$.

Mistake 2: Writing $-\log|1-x|$ or $+\log|1+x|$. …