AP Calculus BC Unit 6: Integration Techniques (Concepts, Methods, and Common Pitfalls)

Finding Antiderivatives and Indefinite Integrals: Basic Rules

What an antiderivative is (and why you care)

An antiderivative of a function f(x) is a function F(x) whose derivative is f(x). In symbols:

F'(x) = f(x)

This matters because integration is the “reverse” of differentiation. When you compute a definite integral (area/accumulated change), you typically do it by first finding an antiderivative and then applying the Fundamental Theorem of Calculus.

Indefinite integrals and the constant of integration

An indefinite integral represents the entire family of antiderivatives:

\int f(x)\,dx = F(x) + C

The +C is crucial: if F'(x)=f(x), then (F(x)+C)'=f(x) for any constant C. A very common source of lost points is forgetting +C on indefinite integrals (but not on definite integrals, where the constant cancels).

Notation connections (same idea, different language)

You’ll see equivalent ways to say the same thing:

Core antiderivative rules (built from derivative rules)

The basic rules are largely reversals of familiar derivative rules.

Linearity is the workhorse: you can integrate term-by-term.

\int \big(af(x)+bg(x)\big)\,dx = a\int f(x)\,dx + b\int g(x)\,dx

Power rule (most common case): for n\neq -1,

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C

The “exception” n=-1 creates the natural log (because \frac{d}{dx}(\ln|x|)=\frac{1}{x}):

\int \frac{1}{x}\,dx = \ln|x| + C

Exponential and trig antiderivatives come from knowing derivative pairs:

\int e^x\,dx = e^x + C

\int a^x\,dx = \frac{a^x}{\ln(a)} + C \quad \text{for } a>0, a\neq 1

\int \cos x\,dx = \sin x + C

\int \sin x\,dx = -\cos x + C

A few especially useful “recognition” forms:

\int \sec^2 x\,dx = \tan x + C

\int \frac{1}{1+x^2}\,dx = \arctan x + C

\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C

Example 1: basic rules with linearity

Compute:

\int \big(3x^2 - 4x + 5\big)\,dx

Use linearity and the power rule:

\int 3x^2\,dx = 3\cdot \frac{x^3}{3} = x^3

\int (-4x)\,dx = -4\cdot \frac{x^2}{2} = -2x^2

\int 5\,dx = 5x

So,

\int \big(3x^2 - 4x + 5\big)\,dx = x^3 - 2x^2 + 5x + C

Example 2: the n=-1 log pattern

Compute:

\int \left(\frac{6}{x} + e^x\right)\,dx

Split it:

\int \frac{6}{x}\,dx = 6\ln|x|

\int e^x\,dx = e^x

So,

\int \left(\frac{6}{x} + e^x\right)\,dx = 6\ln|x| + e^x + C

Exam Focus

• Typical question patterns:
  • “Find the indefinite integral” of a polynomial/trig/exponential expression using linearity.
  • “Find an antiderivative” that satisfies an initial condition such as F(1)=2.
  • Use an antiderivative to evaluate a definite integral via F(b)-F(a).
• Common mistakes:
  • Forgetting +C on indefinite integrals.
  • Using the power rule on \int \frac{1}{x}\,dx (it does not give \frac{x^0}{0}; it gives \ln|x|).
  • Dropping absolute value bars in \ln|x|.

Integrating Using Substitution

What substitution is (reverse chain rule)

Substitution (often called u-substitution) is the main technique for integrals where the integrand contains a “function inside a function.” It is the reverse of the chain rule.

If you recognize something like “a function and (a constant multiple of) its derivative,” substitution turns a complicated integral into a simpler one.

The guiding pattern is:

\int f(g(x))g'(x)\,dx = \int f(u)\,du

where you set:

u=g(x)

and then

du=g'(x)\,dx

Why it matters

Many integrals that look impossible with basic rules become straightforward once you “rename” the inside expression as a single variable. This is especially important in AP Calculus because substitution also supports definite integrals, differential equations, and error checking (differentiating your answer).

How to do it (a reliable process)

1. Choose u as a meaningful “inside” expression (often what’s inside parentheses, radicals, or exponents).
2. Compute du and see if du matches the remaining factors in the integrand (up to a constant multiple).
3. Rewrite the integral completely in terms of u and du.
4. Integrate with respect to u.
5. Substitute back in terms of x.

For definite integrals, you have two valid approaches:

• Substitute back to x and then plug in the original limits.
• Or change the limits to the corresponding u-values and stay in u the whole time.

Example 1: classic “inside function + derivative”

Compute:

\int 2x\cos(x^2)\,dx

The inside is x^2 and its derivative is 2x, which is present.

Let:

u=x^2

Then:

du=2x\,dx

So the integral becomes:

\int \cos(u)\,du = \sin(u) + C

Substitute back:

\sin(x^2) + C

Example 2: definite integral with changed limits

Evaluate:

\int_0^1 \frac{3x^2}{1+x^3}\,dx

Let:

u=1+x^3

Then:

du=3x^2\,dx

Change limits: when x=0, u=1. When x=1, u=2.

So:

\int_0^1 \frac{3x^2}{1+x^3}\,dx = \int_1^2 \frac{1}{u}\,du

Integrate:

\int_1^2 \frac{1}{u}\,du = \left[\ln|u|\right]_1^2 = \ln 2 - \ln 1 = \ln 2

What can go wrong (selection of u)

A common frustration is picking u so that du doesn’t help. For instance, in

\int x\sqrt{1+x^2}\,dx

choosing u=\sqrt{1+x^2} works, but it’s harder than choosing the simpler inside expression u=1+x^2, because du=2x\,dx matches the leftover x\,dx nicely.

Exam Focus

• Typical question patterns:
  • Integrals that match a chain-rule structure, especially with trig, exponential, and rational expressions.
  • Definite integrals where changing bounds makes the algebra cleaner.
  • “Find the average value” problems where substitution simplifies the integral.
• Common mistakes:
  • Forgetting to rewrite every x in terms of u before integrating.
  • Changing variables but not changing limits (when you intended to stay in u).
  • Missing a constant factor in du (for example, needing to multiply/divide by 2).

Integrating Using Long Division and Completing the Square

Why algebra is sometimes the real integration technique

A surprising number of “integration technique” problems are actually algebra problems first. If the integrand is a rational function (a quotient of polynomials), your first job is to rewrite it into forms you already know how to integrate.

Two high-impact algebra moves are:

• Long division (when the numerator’s degree is at least the denominator’s degree).
• Completing the square (to create recognizable forms involving \arctan or logs).

Long division for improper rational functions

A rational function is improper if:

\deg(\text{numerator}) \ge \deg(\text{denominator})

In that case, do polynomial long division to rewrite:

\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}

where S(x) is a polynomial and R(x) has lower degree than Q(x). You can integrate S(x) easily, and the remaining proper rational part may be handled with partial fractions (later in these notes).

Example 1: long division to split the integral

Compute:

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

Divide x^2+1 by x-1:

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

Now integrate term-by-term:

\int \left(x+1+\frac{2}{x-1}\right)dx = \int x\,dx + \int 1\,dx + 2\int \frac{1}{x-1}\,dx

Compute each:

\int x\,dx = \frac{x^2}{2}

\int 1\,dx = x

Let u=x-1 for the log term, or use the known form:

\int \frac{1}{x-1}\,dx = \ln|x-1|

So:

\int \frac{x^2+1}{x-1}\,dx = \frac{x^2}{2} + x + 2\ln|x-1| + C

Completing the square to trigger standard inverse trig/log forms

Completing the square is especially useful when you see a quadratic that doesn’t factor nicely. The goal is to rewrite:

x^2+bx+c = (x+\frac{b}{2})^2 + \left(c-\frac{b^2}{4}\right)

A key recognition: integrals of the form

\int \frac{1}{(x-h)^2+a^2}\,dx

produce an arctangent:

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

Example 2: completing the square to integrate a rational expression

Compute:

\int \frac{1}{x^2+4x+13}\,dx

Complete the square:

x^2+4x+13 = (x+2)^2 + 9

So:

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

Match the arctan pattern with h=-2 and a=3:

\int \frac{1}{(x+2)^2+9}\,dx = \frac{1}{3}\arctan\left(\frac{x+2}{3}\right) + C

What can go wrong (and how to prevent it)

• After long division, students sometimes forget to integrate the polynomial part (they only integrate the remainder).
• When completing the square, small arithmetic mistakes (especially the constant term) derail the whole problem. A good habit is to expand your completed-square form to verify it matches the original quadratic.

Exam Focus

• Typical question patterns:
  • Rational integrals where you must do long division before any other technique applies.
  • Quadratics that don’t factor, pushing you toward completing the square and an \arctan form.
  • Mixed problems where long division creates a simpler rational piece for partial fractions.
• Common mistakes:
  • Skipping long division when \deg(P) \ge \deg(Q).
  • Incorrect completed-square rewriting (wrong constant).
  • Forgetting that \int \frac{1}{(x-h)^2+a^2}dx needs a factor \frac{1}{a} in front of \arctan.

Integration by Parts

What it is (reverse product rule)

Integration by parts is the technique for integrals of products where substitution doesn’t fit well (for example, x e^x, x\sin x, \ln x). It comes directly from the product rule.

Start with the product rule:

(uv)' = u'v + uv'

Integrate both sides:

\int (uv)'\,dx = \int u'v\,dx + \int uv'\,dx

Since \int (uv)'dx = uv, rearrange to get the integration by parts formula:

\int u\,dv = uv - \int v\,du

Here, u is something you choose to differentiate, and dv is the remaining part you integrate.

Why it matters

Integration by parts is one of the few “general-purpose” techniques for products. Many AP problems are designed so that:

• differentiating one factor simplifies it (like \ln x becoming \frac{1}{x}), and
• integrating the other factor is easy (like e^x or \sin x).

How to choose u and dv

You’re trying to make \int v\,du simpler than the original integral.

A helpful heuristic is LIATE for choosing u (not a rule, but a strong hint):

• Logarithmic
• Inverse trig
• Algebraic
• Trig
• Exponential

Choose u earlier in that list when possible.

Example 1: polynomial times exponential

Compute:

\int x e^x\,dx

Choose:

u=x

dv=e^x\,dx

Then:

du=dx

v=e^x

Apply the formula:

\int x e^x\,dx = x e^x - \int e^x\,dx

And:

\int e^x\,dx = e^x

So:

\int x e^x\,dx = x e^x - e^x + C = e^x(x-1)+C

Example 2: integrating \ln x

Compute:

\int \ln x\,dx

This is a classic case where you “create” a product by writing \ln x = 1\cdot \ln x.

Choose:

u=\ln x

dv=1\,dx

Then:

du=\frac{1}{x}\,dx

v=x

Apply integration by parts:

\int \ln x\,dx = x\ln x - \int x\cdot \frac{1}{x}\,dx

Simplify the integrand:

\int x\cdot \frac{1}{x}\,dx = \int 1\,dx = x

So:

\int \ln x\,dx = x\ln x - x + C

What can go wrong

• Picking u and dv so that \int v\,du becomes messier (for instance, choosing u=e^x and dv=x\,dx for \int x e^x dx).
• Forgetting parentheses when subtracting the new integral.
• Dropping constants when integrating or differentiating parts.

Exam Focus

• Typical question patterns:
  • Products like x e^x, x\sin x, x\ln x where substitution doesn’t directly apply.
  • Integrals involving \ln x or inverse trig functions.
  • Sometimes repeated integration by parts (less common, but possible).
• Common mistakes:
  • Using substitution when there is no clear “inside + derivative” match.
  • Incorrectly computing v from dv (especially sign errors with trig).
  • Forgetting that the goal is to simplify the remaining integral \int v\,du.

Integrating Using Linear Partial Fractions

What partial fractions does (turns rational functions into simpler pieces)

Partial fraction decomposition rewrites a rational function as a sum of simpler rational expressions whose antiderivatives you already know (usually logarithms and power functions).

This technique is most straightforward when the denominator factors into linear factors such as (x-a). In AP Calculus BC, you’ll commonly see rational functions where the denominator is already factored or factorable.

When you can (and should) use it

You typically use partial fractions when:

• you have a proper rational function (degree of numerator less than degree of denominator), and
• the denominator factors into linear factors (possibly repeated).

If it’s not proper, do long division first (as in the earlier section).

The basic forms for linear factors

If the denominator includes distinct linear factors, you decompose like this:

\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}

If a linear factor repeats, you need a term for each power up to the repetition:

\frac{P(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}

Each resulting term integrates using log or a power rule.

Example 1: distinct linear factors

Compute:

\int \frac{5x+1}{(x-1)(x+2)}\,dx

Set up the decomposition:

\frac{5x+1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}

Combine the right side:

\frac{A}{x-1} + \frac{B}{x+2} = \frac{A(x+2)+B(x-1)}{(x-1)(x+2)}

So the numerators must match:

A(x+2)+B(x-1) = 5x+1

Expand:

A x + 2A + B x - B = (A+B)x + (2A-B)

Match coefficients with 5x+1:

A+B=5

2A-B=1

Solve: from A+B=5, B=5-A. Substitute:

2A-(5-A)=1

3A-5=1

3A=6

A=2

Then B=3.

Now integrate:

\int \frac{5x+1}{(x-1)(x+2)}\,dx = \int \left(\frac{2}{x-1}+\frac{3}{x+2}\right)dx

So:

\int \frac{2}{x-1}\,dx = 2\ln|x-1|

\int \frac{3}{x+2}\,dx = 3\ln|x+2|

Final answer:

2\ln|x-1| + 3\ln|x+2| + C

Example 2: repeated linear factor

Compute:

\int \frac{1}{(x-1)^2}\,dx

This is already a simple power form:

\frac{1}{(x-1)^2} = (x-1)^{-2}

Let u=x-1 (or apply the shifted power rule carefully). Using substitution:

u=x-1

du=dx

Then:

\int u^{-2}\,du = \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C

Substitute back:

-\frac{1}{x-1} + C

What can go wrong

• Forgetting long division first when the rational function is improper.
• Writing an incomplete decomposition (for repeated factors, you need multiple terms).
• Solving for coefficients incorrectly (a single algebra slip causes a wrong integral).

Exam Focus

• Typical question patterns:
  • “Evaluate \int \frac{P(x)}{Q(x)}dx” where Q(x) factors into linear factors and the result involves logs.
  • Problems that explicitly say “use partial fractions” or implicitly require it after long division.
  • Definite integrals that become sums of logarithms (be careful with absolute value and domain).
• Common mistakes:
  • Missing a term for a repeated factor (for example, using only \frac{A}{x-a} for (x-a)^2).
  • Dropping absolute value bars in log answers.
  • Combining logs incorrectly when simplifying (simplifying is optional; correctness is not).

Evaluating Improper Integrals

What makes an integral “improper”

An improper integral is a definite integral where the usual definition “area from a to b” breaks because:

1. The interval is infinite (like [1,\infty) or (-\infty,2]), or
2. The integrand becomes infinite somewhere in the interval (a vertical asymptote), such as at an endpoint or within [a,b].

Improper integrals are evaluated using limits. The key idea is: you don’t “plug in infinity.” You replace the problematic bound with a variable and take a limit.

Type 1: infinite intervals

For an integral to infinity:

\int_a^{\infty} f(x)\,dx = \lim_{b\to \infty}\int_a^b f(x)\,dx

Similarly:

\int_{-\infty}^{b} f(x)\,dx = \lim_{a\to -\infty}\int_a^b f(x)\,dx

Type 2: infinite discontinuities (vertical asymptotes)

If f(x) blows up at an endpoint a:

\int_a^b f(x)\,dx = \lim_{t\to a^+}\int_t^b f(x)\,dx

If it blows up at an interior point c in [a,b], you must split the integral:

\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx

and evaluate each as a limit. The integral converges only if both pieces converge.

Why it matters

Improper integrals test whether an infinite process produces a finite accumulation. This connects directly to big ideas like infinite series (convergence/divergence), probability density functions over infinite domains, and physical quantities like total mass with density that decays over distance.

Example 1: a convergent improper integral on an infinite interval

Evaluate:

\int_1^{\infty} \frac{1}{x^2}\,dx

Write as a limit:

\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b\to \infty} \int_1^b x^{-2}\,dx

Antiderivative:

\int x^{-2}\,dx = -x^{-1} + C

Evaluate from 1 to b:

\int_1^b x^{-2}\,dx = \left[-\frac{1}{x}\right]_1^b = -\frac{1}{b} - \left(-1\right) = 1-\frac{1}{b}

Take the limit:

\lim_{b\to \infty}\left(1-\frac{1}{b}\right)=1

So the improper integral converges to 1.

Example 2: a divergent improper integral (log growth)

Evaluate:

\int_1^{\infty} \frac{1}{x}\,dx

Set up the limit:

\int_1^{\infty} \frac{1}{x}\,dx = \lim_{b\to \infty}\int_1^b \frac{1}{x}\,dx

Antiderivative:

\int \frac{1}{x}\,dx = \ln|x| + C

Evaluate:

\int_1^b \frac{1}{x}\,dx = \left[\ln x\right]_1^b = \ln b - \ln 1 = \ln b

Now the limit:

\lim_{b\to \infty} \ln b = \infty

So the integral diverges.

Example 3: improper due to an infinite discontinuity at an endpoint

Evaluate:

\int_0^1 \frac{1}{\sqrt{x}}\,dx

The integrand is infinite at x=0. Rewrite using a limit:

\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{t\to 0^+} \int_t^1 x^{-1/2}\,dx

Antiderivative:

\int x^{-1/2}\,dx = 2x^{1/2} + C

Evaluate:

\int_t^1 x^{-1/2}\,dx = \left[2\sqrt{x}\right]_t^1 = 2 - 2\sqrt{t}

Take the limit:

\lim_{t\to 0^+} \left(2 - 2\sqrt{t}\right)=2

So this improper integral converges to 2.

A useful convergence fact (power behavior)

You’ll often compare integrals like \int_1^{\infty} \frac{1}{x^p}dx. This is a standard convergence benchmark:

• It converges if p>1.
• It diverges if p\le 1.

If you use this fact, you should still be able to justify with the limit definition (or show the antiderivative behavior), but recognizing it quickly helps you decide what to do.

Exam Focus

• Typical question patterns:
  • “Determine whether the improper integral converges or diverges. If it converges, find its value.”
  • Improper integrals that require splitting at a discontinuity inside the interval.
  • Problems that combine an integration technique (like substitution) with an improper limit.
• Common mistakes:
  • Treating infinity like a number (plugging in \infty instead of writing a limit).
  • Forgetting to split at an interior asymptote and incorrectly evaluating across the discontinuity.
  • Concluding “convergent” just because an antiderivative exists; the limit behavior is what matters.