Selecting Techniques for Antidifferentiation: Study Guide (Calculus AB 6.14)

Course scope. This topic is shared by Calculus AB and Calculus BC. The basic rules, substitution, long division and completing the square are examinable for both. Integration by parts and linear partial fractions are BC only, and are marked that way below.

A note on notation

This page has no equation renderer. ∫ f(x) dx is an indefinite integral (a family of antiderivatives, so it ends with + C). ∫ (a to b) f(x) dx is the definite integral from x = a to x = b, which is a number. [F(x)] (a to b) means F(b) − F(a). Angles are in radians.

Choosing is the skill

In Topics 6.8 to 6.10 each page told you which technique to use. On a test, nobody tells you. Deciding what kind of integral you are looking at is the whole of Topic 6.14.

There is no new formula here, only a habit: classify first, calculate second. Before you write any working, ask what the integrand is made of. A sum of powers? A function times the derivative of its inside? A fraction whose top has a higher degree than its bottom? Each answer points to a technique.

You did the same with derivatives in Topic 3.5. For antiderivatives the choice matters more: a wrong choice usually gives no answer at all, because the algebra gets worse instead of better.

The toolkit and the signal for each tool

Every technique has a signal: a feature of the integrand that tells you to use it.

Technique	Signal in the integrand	First move	Topic
Basic rule	Matches a known derivative: xⁿ, eˣ, 1/x, sin x, cos x, sec²x, 1/(1 + x²)	Write the antiderivative directly	6.8
Rewrite	Product or quotient you can expand, a fraction over a single term, roots	Expand, split into separate terms, write √x as x^(1/2)	6.8
Substitution	An inner function g(x) together with g′(x), up to a constant factor	Let u = g(x), so du = g′(x) dx	6.9
Long division	Rational function with degree of top ≥ degree of bottom	Divide to get a polynomial plus a proper fraction	6.10
Completing the square	Quadratic in the denominator with no real roots and no matching x on top	Write it as (x − h)² + k², then use arctan	6.10
Integration by parts (BC only)	Product of two unlike types, such as a power times a log, exponential or trig function	∫ u dv = uv − ∫ v du, with u chosen so du is simpler	6.11
Linear partial fractions (BC only)	Proper rational function whose denominator factors into different linear factors	Split into A/(x − p) + B/(x − q), then use ln	6.12

Two habits make the table work:

• Look at the whole integrand, not one part of it. In x/(x² + 9), the x on top is the signal. In 1/(x² + 9) it is missing, and the technique changes.
• Rewrite before you give up. Many integrals that look hard become basic after expanding, splitting or dividing.

A decision path

The path is a guide, not a law. Sometimes two routes work. For ∫ x²(x³ + 2) dx you can expand to ∫ (x⁵ + 2x²) dx or substitute u = x³ + 2. Both are correct, and the two answers differ only by a constant. Choose the shorter route.

Lookalike integrals: small changes, different techniques

The best way to learn the signals is to compare integrals that look almost the same. Every integral below has x² ± 9 in the denominator.

Integral	Signal	Technique	Antiderivative
∫ x/(x² + 9) dx	x on top is half the derivative of x² + 9	Substitution, u = x² + 9	½ ln(x² + 9) + C
∫ 1/(x² + 9) dx	No x on top; x² + 9 has no real roots	arctan form, with 9 = 3²	(1/3) arctan(x/3) + C
∫ x²/(x² + 9) dx	Top degree 2 = bottom degree 2	Long division: 1 − 9/(x² + 9)	x − 3 arctan(x/3) + C
∫ x³/(x² + 9) dx	Top degree 3 > bottom degree 2	Long division: x − 9x/(x² + 9), then substitution	x²/2 − (9/2) ln(x² + 9) + C
∫ 1/(x² − 9) dx (BC only)	x² − 9 = (x − 3)(x + 3), different linear factors	Partial fractions	(1/6) ln

Notice three things.

• Removing the x on top turns a logarithm into an arctangent.
• Changing + 9 to − 9 turns an arctangent into logarithms, because the denominator now factors.
• x² + 9 is always positive, so ln(x² + 9) needs no absolute value bars. ln|x + 2| does.

Worked example 1: classify, then integrate

Question. Find each indefinite integral. For each one, name the signal and the technique before you start.

(a) ∫ (x² − 4)/√x dx (b) ∫ sec²(3x) e^(tan 3x) dx (c) ∫ (x + 5)/(x + 2) dx (d) ∫ 1/(x² − 6x + 13) dx

(a) Signal: a fraction over a single term with a root. Technique: rewrite with powers and split.

1. √x = x^(1/2), so (x² − 4)/√x = x^(3/2) − 4x^(−1/2).
2. Power rule on each term: x^(5/2)/(5/2) − 4 × x^(1/2)/(1/2).
3. ∫ (x² − 4)/√x dx = (2/5)x^(5/2) − 8√x + C

Do not try substitution here. There is no inner function, just a quotient you can split.

(b) Signal: the exponent tan 3x is an inner function, and its derivative 3 sec²(3x) is present apart from the factor 3. Technique: substitution.

1. Let u = tan 3x. Then du = 3 sec²(3x) dx, so sec²(3x) dx = (1/3) du.
2. The integral becomes ∫ (1/3) eᵘ du = (1/3) eᵘ + C.
3. ∫ sec²(3x) e^(tan 3x) dx = (1/3) e^(tan 3x) + C

(c) Signal: a rational function with top degree 1 and bottom degree 1. Technique: long division (or “add and subtract”).

1. Write x + 5 = (x + 2) + 3. So (x + 5)/(x + 2) = 1 + 3/(x + 2).
2. Integrate each term: ∫ 1 dx = x and ∫ 3/(x + 2) dx = 3 ln|x + 2|.
3. ∫ (x + 5)/(x + 2) dx = x + 3 ln|x + 2| + C

(d) Signal: a quadratic denominator with nothing on top. Check for real roots: the discriminant is 36 − 52 = −16 < 0, so it does not factor. Technique: complete the square, then arctan.

1. x² − 6x + 13 = (x² − 6x + 9) + 4 = (x − 3)² + 2².
2. Let u = x − 3, so du = dx. The integral is ∫ 1/(u² + 2²) du = (1/2) arctan(u/2) + C.
3. ∫ 1/(x² − 6x + 13) dx = (1/2) arctan((x − 3)/2) + C

Check (d) by differentiating. The derivative of arctan(v) is v′/(1 + v²). With v = (x − 3)/2, v′ = 1/2. So the derivative of the answer is (1/2) × (1/2) / (1 + (x − 3)²/4) = (1/4) × 4/(4 + (x − 3)²) = 1/((x − 3)² + 4) = 1/(x² − 6x + 13). It matches the integrand.

Worked example 2: split first, then two different techniques

Question. Evaluate ∫ (0 to 2) (x + 2)/(x² − 4x + 8) dx exactly, without a calculator.

1. Classify. The top has degree 1, the bottom degree 2, so no long division. The derivative of the bottom is 2x − 4 = 2(x − 2). The top is x + 2, which is not a constant multiple of x − 2. So substitution alone will not work.

2. Split the top into “a multiple of the derivative” plus “a constant”: x + 2 = (x − 2) + 4. So

   ∫ (0 to 2) (x + 2)/(x² − 4x + 8) dx = ∫ (0 to 2) (x − 2)/(x² − 4x + 8) dx + ∫ (0 to 2) 4/(x² − 4x + 8) dx

3. First piece: substitution. Let u = x² − 4x + 8, so du = (2x − 4) dx and (x − 2) dx = ½ du. Change the limits: x = 0 gives u = 8, and x = 2 gives u = 4. Keep them in that order, even though 8 > 4. ∫ (8 to 4) ½ × (1/u) du = ½ [ln u] (8 to 4) = ½ (ln 4 − ln 8) = −½ ln 2.

4. Second piece: complete the square. x² − 4x + 8 = (x − 2)² + 2². Let v = x − 2, so dv = dx, and v runs from −2 to 0. ∫ (−2 to 0) 4/(v² + 2²) dv = 4 × ½ [arctan(v/2)] (−2 to 0) = 2(0 − (−π/4)) = π/2, since arctan(−1) = −π/4.

5. Add.

   ∫ (0 to 2) (x + 2)/(x² − 4x + 8) dx = π/2 − ½ ln 2

Size check. π/2 ≈ 1.571 and ½ ln 2 ≈ 0.347, so the integral is about 1.224. The integrand rises from 0.25 at x = 0 to 1 at x = 2, and the interval has width 2, so the value must lie between 0.5 and 2. It does.

Common slip. Writing “∫ (x + 2)/(x² − 4x + 8) dx = ½ ln(x² − 4x + 8)” pretends the top is half the derivative of the bottom. It is not, and the answer would be −½ ln 2 ≈ −0.347: negative, for a positive integrand.

When there is no formula

Some integrands have no antiderivative that can be written with familiar functions (powers, roots, exponentials, logarithms, trig and inverse trig functions, combined in finitely many steps). Examples include e^(−x²), sin(x²), cos(x²) and √(1 + x³). No substitution or rewriting will produce a formula, because none exists.

That does not mean the definite integral fails to exist. A continuous function on [a, b] always has a definite integral. You find its value in one of two ways:

• With technology: on a calculator section, ∫ (0 to 1) e^(−x²) dx ≈ 0.747.
• As an accumulation function: F(x) = ∫ (0 to x) e^(−t²) dt is an antiderivative of e^(−x²) by the Fundamental Theorem (Topic 6.4), even though it has no formula in familiar functions.

So “no closed form” is a real possibility. But check first that you have not missed a signal: x e^(−x²) does have a formula, −½ e^(−x²) + C, by substitution.

Worked example 3 (BC only): substitution, parts or partial fractions?

Calculus AB students can skip this example.

Question. Choose a technique for each, then integrate (x > 0 in (a) and (b)).

(a) ∫ (ln x)/x dx (b) ∫ (ln x)/x² dx (c) ∫ 6/(x² + 4x − 5) dx

(a) Signal: ln x is an inner function and 1/x, its derivative, is a factor. Technique: substitution. With u = ln x, du = (1/x) dx: ∫ u du = u²/2 + C, so ∫ (ln x)/x dx = (ln x)²/2 + C.

(b) Signal: the extra power of x means 1/x² is not the derivative of ln x, so substitution fails. It is a product of a logarithm and a power. Technique: integration by parts with u = ln x and dv = x⁻² dx, so du = (1/x) dx and v = −1/x. ∫ (ln x)/x² dx = −(ln x)/x − ∫ (−1/x)(1/x) dx = −(ln x)/x + ∫ x⁻² dx. So ∫ (ln x)/x² dx = −(ln x)/x − 1/x + C.

(c) Signal: a proper rational function, and x² + 4x − 5 = (x − 1)(x + 5) has two different linear factors. Technique: linear partial fractions. 6/((x − 1)(x + 5)) = A/(x − 1) + B/(x + 5). Then 6 = A(x + 5) + B(x − 1). At x = 1: 6 = 6A, so A = 1. At x = −5: 6 = −6B, so B = −1. ∫ 6/(x² + 4x − 5) dx = ln|x − 1| − ln|x + 5| + C, or ln|(x − 1)/(x + 5)| + C.

Parts (a) and (b) differ by one power of x, yet need different techniques.

Always check by differentiating

Differentiate your answer and compare it with the integrand. This catches lost constants, wrong signs and the wrong technique. If two methods give answers that look different, differentiate both: they can differ only by a constant.

Common misconceptions

• “Start with substitution and see what happens.” Substitution only works when the derivative of the inner function is present up to a constant factor. Check for that signal first.
• Dividing by a variable to “fix” du. If du = 2x dx and there is no x in the integrand, you cannot write dx = du/(2x) and carry on. A leftover x means substitution is the wrong choice.
• “Integrate the top and the bottom separately.” ∫ f/g dx is not (∫ f dx)/(∫ g dx). The same goes for products.
• Using ln for every fraction. ∫ 1/(x² + 9) dx is not ln(x² + 9) + C. The ln pattern needs the derivative of the bottom on top.
• Completing the square when the quadratic factors. If the discriminant is positive, the denominator has real roots, and completing the square gives (x − h)² − k², which is not an arctan form. In BC, use partial fractions.
• Skipping long division. If the top’s degree is at least the bottom’s, divide first. Substitution or arctan on a top-heavy fraction leads nowhere.
• Forgetting new limits after substitution. In a definite integral, either change the limits to u-values or switch back to x and use the original x-limits. Never mix the two.
• “Every integral has a formula.” Some, such as ∫ e^(−x²) dx, do not. Use technology or an accumulation function for a definite value.
• Dropping + C or the absolute value. Write ln|x + 2|, not ln(x + 2), unless the expression inside is always positive.

Where this leads

Topic 6.14 closes the antidifferentiation part of Unit 6. In Unit 7 you will solve differential equations by separating variables (Topic 7.6); each side of the separated equation is an integral that needs exactly this selection skill. The next topic, Topic 7.1, Modeling Situations with Differential Equations, starts that work. If a technique felt shaky, go back to Topic 6.8 (basic rules), Topic 6.9 (substitution) or Topic 6.10 (long division and completing the square). BC students can also revisit Topic 6.11 (integration by parts) and Topic 6.12 (linear partial fractions). 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.