Q.Integrate the following function: $\frac{x^3 + x + 1}{x^2 - 1}$

Idea: the fraction is improper (numerator degree $3$ > denominator degree $2$), so divide first, then use partial fractions on the remainder.

Divide: $x^3+x+1=x(x^2-1)+(2x+1)$, so

Decompose the proper part with $x^2-1=(x-1)(x+1)$:

• $x=1$: $3=2A\Rightarrow A=\tfrac{3}{2}$
• $x=-1$: $-1=-2B\Rightarrow B=\tfrac{1}{2}$ …

The first move: long division

Whenever the numerator's degree is at least the denominator's, partial fractions cannot start — you must divide first. Here numerator $x^3+x+1$ has degree $3$ and denominator $x^2-1$ has degree $2$, so we divide.

Subtract: $(x^3+x+1)-(x^3-x)=2x+1$. So the quotient is $x$ and the remainder is $2x+1$:

Partial fractions on the remainder

Factor $x^2-1=(x-1)(x+1)$ and write

Clear denominators: $2x+1=A(x+1)+B(x-1)$.

• $x=1$: $\;3=2A\Rightarrow A=\dfrac{3}{2}$
• $x=-1$: $\;-1=-2B\Rightarrow B=\dfrac{1}{2}$

So

Integrate term by term …

Method: Divide First, then Partial Fractions (Improper Fractions)

Use this whenever the numerator's degree is greater than or equal to the denominator's — partial fractions cannot start until the fraction is made proper.

Steps

Step 1: Compare degrees.

If $\deg(\text{numerator}) \ge \deg(\text{denominator})$, the fraction is improper; do polynomial (long) division first.

Step 2: Divide to get quotient + proper remainder.

For example $\dfrac{x^3+x+1}{x^2-1} = x + \dfrac{2x+1}{x^2-1}$. …

Common Mistakes

Mistake 1: Jumping straight to partial fractions on an improper fraction.

Why it's wrong: partial-fraction decomposition is only valid when the numerator degree is strictly less than the denominator degree; here $\deg=3>\deg=2$, so the setup fails. Correct approach: do long division first, getting $x+\dfrac{2x+1}{x^2-1}$.

Mistake 2: Forgetting to integrate the quotient term. …