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

The key idea is that the numerator’s degree is higher than the denominator’s, so we first perform polynomial long division.

Divide $x^3 - x^2 + x - 1$ by $x - 1$:

1. $x^3 \div x = x^2$, multiply back: $x^3 - x^2$, subtract → remainder $0 + x - 1$.
2. $x \div x = 1$, multiply back: $x - 1$, subtract → remainder $0$. …

When you see a rational function where the numerator’s degree (3) is greater than the denominator’s degree (1), your first instinct should be: divide. The denominator $x-1$ is linear, so the division is quick — and it removes the fraction entirely. The result will be a plain polynomial, which you can integrate term by term.

Why does this work? The expression $\frac{x^3 - x^2 + x - 1}{x-1}$ is just a fraction. If you can rewrite it as $(\text{polynomial}) + \frac{\text{remainder}}{x-1}$, and if the remainder turns out to be zero, you’ve eliminated the denominator. That’s exactly what happens here — the numerator is divisible by $x-1$, so the quotient is a clean quadratic.

Let’s do the division step by step.

1. Set up the long division.

   Divide $x^3 - x^2 + x - 1$ by $x - 1$.

   Ask: what do I multiply $(x-1)$ by to get the leading term $x^3$? The answer is $x^2$, because $x^2 \cdot (x-1) = x^3 - x^2$.

2. Subtract and bring down the next term.

   Subtract $(x^3 - x^2)$ from the numerator:

The $x^3$ and $x^2$ terms cancel completely. Now bring down the remaining $+x - 1$.

3. Repeat with the new polynomial $x - 1$. What multiplies $(x-1)$ to give $x$? The answer is $1$, because $1 \cdot (x-1) = x - 1$. Subtract: $(x - 1) - (x - 1) = 0$. The remainder is zero.

So the division yields:

$\frac{x^3 - x^2 + x - 1}{x - 1} = x^2 + 1.$ …

Method: Divide (or factor) an improper rational integrand

When the numerator has higher degree than a linear denominator, long division — or spotting a factor of the denominator in the numerator — clears the fraction.

Steps

Step 1: Confirm it is improper and divide.

$\deg 3>\deg 1$, so divide the cubic by $x-1$; here the remainder is $0$ and the quotient is $x^2+1$. …

Common Mistakes

Mistake 1: Splitting the fraction into $\int\frac{x^3}{x-1}-\int\frac{x^2}{x-1}+\cdots$.

Why it's wrong: each piece is still improper and drags in messy logarithms. Correct approach: divide the whole numerator by $x-1$ once, giving the clean quotient $x^2+1$.

Mistake 2: Assuming a leftover logarithm must appear. …