6.10 Integrating Functions Using Long Division and Completing the Square

AP Calc 6.10 Integration Summary

When an integral has a rational function where the top has a degree equal to or higher than the bottom, use polynomial long division to rewrite it as a polynomial plus a simpler fraction. When the integrand has a quadratic that is not a perfect square, complete the square to turn it into a form that leads to an arctan or arcsin antiderivative. Both moves rearrange the integrand into something you already know how to integrate.

Why This Matters for the AP Calculus Exam

Long division and completing the square are rewriting tools. They do not integrate anything by themselves. They take an integrand that looks impossible and turn it into a sum of pieces you can handle with basic antiderivative rules, u-substitution, or the standard arctan and arcsin forms.

On the AP Calculus exam, these techniques show up most often in multiple-choice questions. A quick scan of the answer choices is a strong clue: if you see options that combine a polynomial term with a natural log term, you are probably looking at a long division problem. If you see arctan or arcsin in the choices, completing the square is likely the path. Both AB and BC students are responsible for this topic. Knowing when to reach for each technique is just as important as the algebra itself.

Key Takeaways

• Use long division when the numerator's degree is greater than or equal to the denominator's degree. This turns an improper rational function into a polynomial plus a proper fraction.
• After dividing, integrate the polynomial part with the power rule and handle the leftover fraction with basic rules, u-substitution, or a log antiderivative.
• Complete the square when the integrand has a quadratic with no obvious perfect square, especially when you expect an inverse trig result.
• Rewrite the quadratic as $(x-h)^2 + k$ or similar, then match it to the standard form $\int \frac{1}{x^2+a^2}dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)+C$.
• Always include $+C$ for indefinite integrals, and for definite integrals adjust your limits if you substitute.
• Watch the answer choices for clues: polynomial plus log signals long division, while arctan or arcsin signals completing the square.

Integrating Using Long Division

A rational function has a polynomial in both the numerator and the denominator. If the numerator's degree is equal to or higher than the denominator's degree, long division is usually the way to go. Dividing first breaks the fraction into a polynomial you can integrate directly plus a smaller fraction that is easier to handle.

Long Division Walkthrough

Evaluate:

$$∫\frac{2x^2-4}{x+1}dx$$

The numerator is a quadratic and the denominator is linear, so the numerator's degree is higher. That is your signal to divide.

Carrying out the polynomial long division of $2x^2-4$ by $x+1$ gives a quotient of $2x - 2$ with a remainder of $-2$, so:

$$\frac{2x^2-4}{x+1} = 2x - 2 - \frac{2}{x+1}$$

Now replace the integrand with the divided form:

$$∫\left(2x-2-\frac{2}{x+1}\right)dx$$

Integrate term by term. The polynomial part uses the power rule, and the last term gives a natural log:

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

Notice the structure of the answer: a polynomial followed by a log term. That pattern is exactly what you look for in the multiple-choice answer choices.

Integrating by Completing the Square

When you see a quadratic expression in the denominator that is not already a perfect square, completing the square can rewrite it into a usable form. You add and subtract the right value to create a perfect square plus a constant. This shows up often when the result is an inverse trig function like arctan or arcsin.

Completing the Square Walkthrough

Evaluate:

$$∫\frac{4}{t^2-4t+20}dt$$

Complete the square on the denominator. You want the form $(t-h)^2+k$:

$$t^2-4t+20=(t^2-4t+4)+20-4$$ $$=(t-2)^2+16$$

Substitute it back into the integral:

$$∫\frac{4}{(t-2)^2+16}dt$$

Pull the constant out front so the numerator is 1:

$$4∫\frac{1}{(t-2)^2+16}dt$$

This now matches the standard arctan form:

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

Here $x = t-2$ and $a = 4$ (since $16 = 4^2$). Apply the formula and keep the constant 4 you factored out:

$$4\cdot\frac{1}{4}\arctan\left(\frac{t-2}{4}\right)+C$$ $$\arctan\left(\frac{t-2}{4}\right)+C$$

The factor of 4 outside and the $\frac{1}{4}$ from the formula cancel cleanly, which is a nice check that you set things up right.

How to Use This on the AP Calculus Exam

MCQ

This topic appears most often in multiple choice. Use the answer choices as a roadmap:

• Choices that mix a polynomial with a natural log term point to long division.
• Choices with arctan or arcsin point to completing the square.

Before grinding through algebra, glance at the form of the answers to decide which technique fits.

Problem Solving

A reliable order of operations:

1. Check the degrees. If the numerator's degree is greater than or equal to the denominator's, divide first.
2. After dividing, integrate the polynomial with the power rule and deal with the remaining fraction separately.
3. If you have a quadratic that is not a perfect square, complete the square and look for an inverse trig match.
4. Match your rewritten integral to a known form, such as the arctan or arcsin antiderivative.
5. Add $+C$ for indefinite integrals. For definite integrals, either change your limits when you substitute or convert back to the original variable before plugging in.

Common Trap

Forgetting to divide when the rational function is improper. If the top degree is at least as big as the bottom degree, you cannot jump straight to a log or arctan. Divide first, then integrate the pieces.

Practice Problems

Practice 1

Evaluate:

$$∫\frac{2x^3+3x^2-17x-27}{x^2-9}dx$$

The numerator degree is higher, so use long division. Dividing gives:

$$∫\left(2x+3+\frac{x}{x^2-9}\right)dx$$

The leftover fraction integrates with a u-substitution since the numerator is a constant multiple of the derivative of $x^2-9$:

$$x^2+3x+\frac{1}{2}\ln(|x^2-9|)+C$$

Practice 2

Evaluate:

$$∫\frac{1}{\sqrt{3-x^2-2x}}dx$$

Complete the square inside the root:

$$-x^2-2x+3=-(x+1)^2+4$$ $$∫\frac{1}{\sqrt{-(x+1)^2+4}}dx$$

This matches the arcsin form, giving:

$$\arcsin\left(\frac{1}{2}(x+1)\right)+C$$

Practice 3

Evaluate:

$$∫\frac{x^2}{x+1}dx$$

The numerator degree is higher, so divide:

$$∫\left(x-1+\frac{1}{x+1}\right)dx$$

Integrate each piece:

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

Common Misconceptions

• Skipping long division on improper fractions. If the numerator's degree is equal to or greater than the denominator's, you must divide before trying to integrate. You cannot treat an improper rational function as if it were already a simple log or arctan.
• Completing the square incorrectly. When you add a value inside the square, you have to subtract the same value (or account for any sign change) so the expression stays equal. In Practice 2, the negative sign in front flips how you balance the constant.
• Mixing up the arctan constant. In the form $\int \frac{1}{x^2+a^2}dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)+C$, the value of $a$ is the square root of the constant, not the constant itself. With $(t-2)^2+16$, you have $a=4$, not $a=16$.
• Dropping the constant of integration. Indefinite integrals always need $+C$. It is an easy point to lose on free response work.
• Forgetting to adjust definite integral limits. If you substitute variables in a definite integral, either change the limits to match the new variable or convert back before evaluating.

Related AP Calculus Guides

• Unit 6 Overview: Integration and Accumulation of Change
• 6.11 Integrating Using Integration by Parts
• 6.1 Integration and Accumulation of Change
• 6.12 Integrating Using Linear Partial Fractions
• 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
• 6.4 The Fundamental Theorem of Calculus and Accumulation Functions