Topic 6.10 Notes – Integrating Functions Using Long Division and Completing the Square

Using Polynomial Long Division

You use long division when the integrand is a rational function
$$\frac{P(x)}{D(x)}$$
and the degree of the numerator is greater than or equal to the degree of the denominator.

That’s your signal that the fraction can be rewritten as:

$$\text{polynomial} + \frac{\text{remainder}}{D(x)}$$

Once it’s split, the integral becomes manageable.

What This Looks Like in Practice

Suppose you’re integrating:

$$\int \frac{x^3 + 2x^2 - 1}{x - 1} \, dx$$

The numerator has degree 3 and the denominator degree 1, so divide. Here’s the setup:

Polynomial long division of $(x^3 + 2x^2 - 1) \div (x - 1)$

After dividing, you get:

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

Now integrate term by term:

• $\int x^2 dx = \frac{x^3}{3}$
• $\int 3x dx = \frac{3x^2}{2}$
• $\int 3 dx = 3x$
• $\int \frac{2}{x-1} dx = 2\ln|x-1|$

Final answer:

$$\frac{x^3}{3} + \frac{3x^2}{2} + 3x + 2\ln|x-1| + C$$

Notice how the leftover fraction almost always turns into a natural log term.

Things That Cost Points

• Forgetting to divide when degrees are equal.
• Dropping the absolute value in $\ln|x-1|$.
• Making small algebra mistakes during division.
• Trying substitution before simplifying.

On multiple-choice, if answers look like “polynomial + ln term,” that’s often your clue long division was needed.

Completing the Square

This shows up when you see a quadratic that isn’t factorable or a perfect square, especially inside:

• A denominator
• A square root
• Something that resembles an inverse trig pattern

The goal is to rewrite the quadratic so it matches a known inverse trig form.

The Two Forms You Want

From your formula sheet:

$$\int \frac{1}{x^2 + a^2} dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C$$

$$\int \frac{1}{\sqrt{a^2 - x^2}} dx = \arcsin\left(\frac{x}{a}\right) + C$$

So you’re trying to turn a messy quadratic into:

• $(x-h)^2 + a^2$ → arctan
• $a^2 - (x-h)^2$ → arcsin

Example with Arctan

Consider:

$$\int \frac{1}{x^2 - 6x + 13} dx$$

Here’s the algebra visually, since this is where most mistakes happen:

Completing the square for $x^2 - 6x + 13$

Half of −6 is −3. Square it to get 9.

$$x^2 - 6x + 13 = (x^2 - 6x + 9) + 4 = (x - 3)^2 + 4$$

Now the integral becomes:

$$\int \frac{1}{(x-3)^2 + 2^2} dx$$

This matches the arctan formula with:

• $a = 2$
• shift $x-3$

So the answer is:

$$\frac{1}{2}\arctan\left(\frac{x-3}{2}\right) + C$$

Definite Integrals

For definite integrals:

• Do the algebra first.
• Integrate fully.
• Plug in bounds carefully.

No $+C$.

With inverse trig answers, use parentheses when evaluating bounds. Calculator sections often include these because evaluating arctan or arcsin numerically is expected.

Strategy Pattern You Should See

When stuck, ask:

1. Is this a rational function with numerator degree ≥ denominator?
   → Long division.
2. Is there a quadratic that won’t factor nicely?
   → Complete the square.
3. Does the result resemble $x^2 + a^2$ or $a^2 - x^2$?
   → Inverse trig.

The calculus part is familiar. The rearranging is the skill.