Integrating Functions Using Long Division and Completing the Square: Study Guide (Calculus AB 6.10)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer, so integrals are written in a compact form. ∫ f(x) dx is an indefinite integral (add + C). ∫ (a to b) f(x) dx is the definite integral from x = a to x = b. [F(x)] (a to b) means F(b) − F(a). arctan and arcsin are the inverse tangent and inverse sine; you may also see them written tan⁻¹ and sin⁻¹.

Why rewrite before integrating?

Some integrands match no basic rule and no obvious substitution as written. Try ∫ (x² + 3x + 1)/(x + 1) dx. Let u = x + 1? Then du = dx, but you are left with (x² + 3x + 1)/u, which still needs work. There is no product or chain-rule pattern to undo.

The fix is to write the same function in a different form that does match the rules you know. Two algebra tools do most of the work:

What you see	First step	What you get
Rational function, degree of top ≥ degree of bottom	Long division	Polynomial + (remainder)/(divisor)
1/(quadratic) where the quadratic has no real roots	Complete the square	an arctan form
1/√(quadratic) with a negative x² term	Complete the square	an arcsin form
(linear)/(quadratic with no real roots)	Split the numerator, then complete the square	a ln term plus an arctan term

The rewriting is pure algebra. It changes how the function looks, not its values, so it changes nothing about the integral.

Where AB stops. If the denominator factors into distinct linear factors, such as x² − 4 = (x − 2)(x + 2), the tool you need is partial fractions. That is a BC-only topic (6.12). In this topic, after division the denominator is usually linear or a quadratic with no real roots. (A perfect square such as (x + 1)² is fine too: ∫ dx/(x + 1)² is a power-rule substitution.)

Tool 1: long division

When the top’s degree is at least the bottom’s, divide, exactly as with numbers. 17/5 = 3 + 2/5, and in the same way:

(x² + 3x + 1)/(x + 1) = x + 2 − 1/(x + 1)

The division, step by step:

1. x² ÷ x = x. Multiply: x(x + 1) = x² + x. Subtract: (x² + 3x + 1) − (x² + x) = 2x + 1.
2. 2x ÷ x = 2. Multiply: 2(x + 1) = 2x + 2. Subtract: (2x + 1) − (2x + 2) = −1.
3. Quotient x + 2, remainder −1. So the fraction equals x + 2 + (−1)/(x + 1).

Check by multiplying back: (x + 2)(x + 1) − 1 = x² + 3x + 2 − 1 = x² + 3x + 1. Correct.

Now each piece is easy: x and 2 are powers, and ∫ 1/(x + 1) dx = ln|x + 1| + C. The graph below shows what the division means.

Two quick cases. When the degrees are equal, the quotient is a constant: (x² + 2)/(x² + 1) = 1 + 1/(x² + 1). When the top is a simple shift of the bottom, you can “add and subtract” instead of dividing: x/(x + 1) = (x + 1 − 1)/(x + 1) = 1 − 1/(x + 1).

Tool 2: completing the square

The target forms are two results you can derive from Topic 6.8 with a substitution (Topic 6.9). For a constant a > 0:

Arctan form. ∫ du/(a² + u²) = (1/a) arctan(u/a) + C

Arcsin form. ∫ du/√(a² − u²) = arcsin(u/a) + C

Where the 1/a comes from. In ∫ du/(a² + u²), take a² out of the bottom: (1/a²) ∫ du/(1 + (u/a)²). Let w = u/a, so du = a dw. This gives (1/a²) · a ∫ dw/(1 + w²) = (1/a) arctan w + C. In the arcsin form, the factor a from du cancels the a from the square root, so no 1/a appears.

To reach these forms, complete the square:

x² + bx + c = (x + b/2)² + (c − b²/4)

• x² + 6x + 13 = (x + 3)² + 4, so a = 2 and u = x + 3. Then ∫ dx/(x² + 6x + 13) = (1/2) arctan((x + 3)/2) + C.
• 6x − x² = −(x² − 6x) = −((x − 3)² − 9) = 9 − (x − 3)², so a = 3 and u = x − 3. Then ∫ dx/√(6x − x²) = arcsin((x − 3)/3) + C.

When does arctan apply? Only if the quadratic has no real roots (discriminant b² − 4ac < 0). Then the constant left after completing the square is positive, and you get a sum of squares. x² − 4 has discriminant 16 > 0, so it is a difference of squares, and arctan does not apply.

Worked example 1: long division in a definite integral

Question. Evaluate ∫ (0 to 2) (x² + 3x + 1)/(x + 1) dx exactly.

1. Compare degrees. Top degree 2, bottom degree 1. Divide first.
2. Divide (shown above): the integrand equals x + 2 − 1/(x + 1).
3. Integrate each term: an antiderivative is x²/2 + 2x − ln|x + 1|. On [0, 2], x + 1 > 0, so write ln(x + 1).
4. Evaluate. At x = 2: 2 + 4 − ln 3 = 6 − ln 3. At x = 0: 0 + 0 − ln 1 = 0.
5. Answer: 6 − ln 3, about 4.90.

Check with the picture. Figure 1 gives the same answer: trapezoid area 6 minus gap area ln 3. The curve runs from height 1 to height 11/3 over a width of 2, so the area must lie between 2 and about 7.3. The answer fits.

Worked example 2: completing the square to reach arctan

Question. Evaluate ∫ (1 to 3) 1/(x² − 2x + 5) dx exactly.

1. Compare degrees. Top degree 0, bottom degree 2. No division needed.
2. Check for real roots. Discriminant (−2)² − 4(5) = −16 < 0. No real roots, so aim for arctan.
3. Complete the square: x² − 2x + 5 = (x − 1)² + 4. So a = 2, and let u = x − 1, with du = dx.
4. Change the limits: x = 1 gives u = 0; x = 3 gives u = 2.
5. Integrate: ∫ (0 to 2) du/(4 + u²) = [(1/2) arctan(u/2)] (0 to 2) = (1/2)(arctan 1 − arctan 0).
6. Evaluate: arctan 1 = π/4 and arctan 0 = 0, so the answer is (1/2)(π/4) = π/8, about 0.393.

Check. The integrand falls from 1/4 at x = 1 to 1/8 at x = 3, over a width of 2. So the area lies between 2 × 1/8 = 0.25 and 2 × 1/4 = 0.5. The answer 0.393 fits.

Worked example 3: splitting the numerator

Question. Find ∫ (x + 4)/(x² + 4x + 8) dx.

1. Compare degrees. Top degree 1, bottom degree 2. No division.
2. Look at the derivative of the bottom: d/dx (x² + 4x + 8) = 2x + 4.
3. Split the top into a multiple of 2x + 4 plus a constant: x + 4 = (1/2)(2x + 4) + 2.
4. Write two integrals: (1/2) ∫ (2x + 4)/(x² + 4x + 8) dx + 2 ∫ 1/(x² + 4x + 8) dx
5. First integral: derivative over function (u = x² + 4x + 8), so it gives (1/2) ln(x² + 4x + 8). The quadratic is always positive (discriminant 16 − 32 < 0), so no absolute value is needed.
6. Second integral: complete the square: x² + 4x + 8 = (x + 2)² + 4, so a = 2. It gives 2 · (1/2) arctan((x + 2)/2) = arctan((x + 2)/2).
7. Answer: ∫ (x + 4)/(x² + 4x + 8) dx = (1/2) ln(x² + 4x + 8) + arctan((x + 2)/2) + C

Check. Differentiate: (1/2)(2x + 4)/(x² + 4x + 8) + (1/2)/(1 + (x + 2)²/4). The second term equals 2/(4 + (x + 2)²) = 2/(x² + 4x + 8). Adding gives (x + 2 + 2)/(x² + 4x + 8) = (x + 4)/(x² + 4x + 8). Correct.

Common misconceptions

• Integrating top and bottom separately. ∫ p(x)/q(x) dx is not (∫ p dx)/(∫ q dx). Rewrite first.
• Forgetting the remainder. After dividing, the remainder term is usually where the ln comes from. Dropping it loses part of the answer.
• Completing the square when the top is too big. If the top’s degree is at least the bottom’s, divide first.
• Writing ln of the quadratic for 1/(quadratic). ∫ 1/(x² + 6x + 13) dx is not ln(x² + 6x + 13). That would need 2x + 6 on top.
• Using 1/a² or a² instead of 1/a and a. From (x − 1)² + 4, a = 2, not 4.
• Sign slips in completing the square. x² − 2x + 5 = (x − 1)² + 4, not (x + 1)² + 4. Expand your square to check.
• Forcing arctan onto a quadratic that factors. 1/(x² − 4) is not an arctan form. It needs partial fractions, which is BC-only.

Where this leads

Topic 6.14, Selecting Techniques for Antidifferentiation, asks you to choose between basic rules, substitution and the rearrangements on this page. BC students add integration by parts and partial fractions before that. Rewritten integrands come up again in Unit 7 (separable differential equations) and Unit 8 (areas and volumes). Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.