Unit 6 Integration Tools: Building Antiderivatives and Choosing Techniques

Finding Antiderivatives and Indefinite Integrals: Basic Rules

What an antiderivative is (and why you should care)

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

F'(x) = f(x)

This matters because differentiation and integration are “inverse” ideas: differentiation tells you an instantaneous rate of change, while integration lets you rebuild a function from its rate of change (up to an unknown starting value). In AP Calculus AB, you constantly move back and forth between these perspectives—for example, when a problem gives you a velocity function and asks for position, or gives you f'(x) and asks for f(x).

An indefinite integral is the notation used for “the set of all antiderivatives”:

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

The constant C is the **constant of integration**. It’s there because if F'(x) = f(x), then \big(F(x) + C\big)' = f(x) for any constant C. In other words, there isn’t one antiderivative—there’s a whole family of them, all vertically shifted.

A helpful analogy: if differentiation is like taking a “slope snapshot” at every point, then antidifferentiation is like reconstructing the original height function from those slope snapshots. But because you don’t know the absolute starting height, you get a family of possible height functions differing by a constant.

Notation you’ll see (and what it means)

You’ll encounter several equivalent ways to express antiderivatives:

A common early mistake is treating dx as “just decoration.” In basic rules it often feels that way, but later (especially with substitution) it becomes a signal for what variable you’re integrating with respect to and how to rewrite the integrand.

Core antiderivative rules (how they work)

The basic rules come directly from reversing derivative rules.

1) Power rule (reverse of the derivative power rule)

If n is a real number with n \neq -1, then:

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

Why the restriction n \neq -1? Because when n=-1, the formula would require dividing by zero. And indeed, x^{-1} = \frac{1}{x} has a different antiderivative (a logarithm).

Example (basic power rule):

Find \int \big(3x^2 - 5x^{1/2}\big)\,dx.

Step 1: Split the integral (linearity):

\int \big(3x^2 - 5x^{1/2}\big)\,dx = \int 3x^2\,dx - \int 5x^{1/2}\,dx

Step 2: Pull out constants:

= 3\int x^2\,dx - 5\int x^{1/2}\,dx

Step 3: Apply the power rule:

= 3\cdot \frac{x^3}{3} - 5\cdot \frac{x^{3/2}}{3/2} + C

Step 4: Simplify:

= x^3 - \frac{10}{3}x^{3/2} + C

Common “what goes wrong” here: students often forget that dividing by 3/2 is multiplying by 2/3.

2) Log rule (the special case for 1/x)

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

The absolute value matters because the derivative of \ln|x| is 1/x for both positive and negative x (excluding x=0). On many AP questions, if you omit absolute values you may lose accuracy unless the context guarantees x>0.

Example (log rule in disguise):

Compute \int \frac{7}{x}\,dx.

Pull out the constant:

\int \frac{7}{x}\,dx = 7\int \frac{1}{x}\,dx = 7\ln|x| + C

3) Linearity rules (why they’re powerful)

Linearity lets you break complicated expressions into simpler pieces:

\int \big(f(x) + g(x)\big)\,dx = \int f(x)\,dx + \int g(x)\,dx

\int kf(x)\,dx = k\int f(x)\,dx

This matters because a lot of “integration technique” in AB is really algebra first: simplify the integrand into a sum of standard forms.

4) Exponential and trig antiderivatives you’re expected to know

These come from reversing common derivative facts:

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

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

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

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

Two very common sign mistakes happen with trig:

• Forgetting that the derivative of \cos(x) is -\sin(x) (so the antiderivative of \sin(x) must be -\cos(x)).
• Dropping the negative sign when integrating -\sin(x).

Example (mixing types):

Find \int \big(2e^x + 4\cos(x) - \frac{3}{x}\big)\,dx.

Integrate term-by-term:

\int 2e^x\,dx = 2e^x

\int 4\cos(x)\,dx = 4\sin(x)

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

So the result is:

2e^x + 4\sin(x) - 3\ln|x| + C

Antiderivatives and initial conditions (how you get a unique function)

Because indefinite integrals produce families, AP problems often give one extra piece of information to pin down C.

If you know F'(x)=f(x) and also F(a)=b, you can find C.

Example (finding the constant):

Given F'(x)=6x-4 and F(2)=5, find F(x).

1) Antidifferentiate:

F(x)=\int (6x-4)\,dx = 3x^2 - 4x + C

2) Use the condition F(2)=5:

5 = 3(2)^2 - 4(2) + C = 12 - 8 + C = 4 + C

So C=1 and:

F(x)=3x^2-4x+1

A frequent mistake is trying to “plug in” before integrating. The condition is about the original function, not its derivative.

Exam Focus

• Typical question patterns:
  • “Find the most general antiderivative” (indefinite integral) of a given algebraic/trig/exponential expression.
  • “Given f'(x) and a point on f, find f(x)” (solve for C).
  • “Integrate a sum by rewriting in standard forms” (e.g., rewrite radicals as powers, rewrite as constants times standard functions).
• Common mistakes:
  • Forgetting +C on indefinite integrals (it signals a family of answers).
  • Misusing the power rule when the exponent is -1 (you must use \ln|x| instead).
  • Sign errors with trig antiderivatives, especially \int \sin(x)\,dx.

Integrating Using Substitution

The big idea: reverse the chain rule

Substitution (often called u-substitution) is the integration technique that reverses the chain rule. The chain rule says:

\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)

So when you see an integrand shaped like “something” times the derivative of that something, substitution helps you collapse it into a single-variable integral.

A classic pattern is:

\int f(g(x))g'(x)\,dx

Let:

u=g(x)

Then:

du=g'(x)\,dx

and the integral becomes:

\int f(u)\,du

Why this matters in AB: many integrals that look complicated are actually simple “inside-out” once you recognize the inner function and its derivative nearby.

How to do substitution reliably (a step-by-step process)

1) Choose u to be an “inside” expression—often what’s inside parentheses, a radical, an exponent, or a denominator.
2) Differentiate to find du.
3) Rewrite the entire integral in terms of u and du (this is the step students skip, and it’s where most errors happen).
4) Integrate with respect to u.
5) Substitute back to return to the original variable.

A good self-check: after rewriting, there should be no leftover x’s anywhere before you integrate in u.

Example 1: linear inner function (the “almost basic” case)

Evaluate:

\int 2x(x^2+5)^7\,dx

Why substitution applies: the integrand contains (x^2+5)^7 and also 2x, which is the derivative of x^2+5.

Step 1: Choose u:

u=x^2+5

Step 2: Differentiate:

du=2x\,dx

Step 3: Rewrite:

\int 2x(x^2+5)^7\,dx = \int u^7\,du

Step 4: Integrate:

\int u^7\,du = \frac{u^8}{8}+C

Step 5: Substitute back:

\frac{(x^2+5)^8}{8}+C

Common pitfall: writing \frac{(x^2+5)^8}{8} but forgetting the constant of integration (still needed).

Example 2: exponential inner function

Evaluate:

\int e^{3x-1}\,dx

Here the “inside” is 3x-1. The derivative is 3, which is almost present (you can fix that by factoring).

Let:

u=3x-1

Then:

du=3\,dx

So:

dx=\frac{1}{3}du

Rewrite and integrate:

\int e^{3x-1}\,dx = \int e^u\cdot \frac{1}{3}du = \frac{1}{3}e^u + C = \frac{1}{3}e^{3x-1}+C

A frequent mistake is to write e^{3x-1}/(3x-1), confusing substitution with a “divide by the inside” trick. Substitution doesn’t work that way—you multiply by the correct differential factor.

Example 3: rational form leading to a logarithm

Evaluate:

\int \frac{5x^4}{x^5+2}\,dx

The denominator x^5+2 has derivative 5x^4, which is exactly the numerator. This is a perfect substitution-for-log pattern.

Let:

u=x^5+2

Then:

du=5x^4\,dx

So:

\int \frac{5x^4}{x^5+2}\,dx = \int \frac{1}{u}\,du = \ln|u| + C = \ln|x^5+2| + C

Why the absolute value is still there: x^5+2 could be negative for some x, so the most general antiderivative uses \ln|\cdot|.

Recognizing when substitution is (and isn’t) the right tool

Substitution is best when:

• You can identify an “inner function” and something proportional to its derivative.
• The integral involves a composition (power of something, exponential of something, trig of something, reciprocal of something).

Substitution is usually not the right tool when:

• The expression is a pure rational function that needs algebraic manipulation first (often long division).
• The integrand is a sum where only one term matches a derivative pattern (you may need to split and handle pieces differently).

A practical habit: before committing to substitution, quickly ask “If I let u be the complicated chunk, do I see du (up to a constant)?”

Exam Focus

• Typical question patterns:
  • “Evaluate an indefinite integral” where the integrand has a clear inner function and its derivative factor (polynomial, exponential, trig, rational).
  • “Find an antiderivative” as part of solving an initial value problem (substitution first, then use the given point to find C).
  • “Rewrite and integrate” problems where you must factor out a constant to make du appear.
• Common mistakes:
  • Choosing u but not converting the entire integral to u (leaving stray x terms).
  • Forgetting the constant multiple adjustment (e.g., missing the factor 1/3 when du=3dx).
  • Treating substitution like a memorized trick rather than a chain-rule reversal—leading to incorrect choices of u.

Integrating Using Long Division and Completing the Square

Why algebraic rewriting is an “integration technique”

Some integrals look hard not because you need a new calculus idea, but because the expression is in an inconvenient algebraic form. In AP Calculus AB, two especially useful algebra moves are:

• Long division for rational functions when the numerator degree is greater than or equal to the denominator degree.
• Completing the square to convert a quadratic expression into a form that matches standard antiderivatives (often involving \ln|x| or \arctan(x)).

The mindset: integration often starts with simplifying the integrand into a sum of pieces you already know how to integrate.

Integrating rational functions using long division

A rational function is a ratio of polynomials:

\frac{P(x)}{Q(x)}

If \deg(P) \ge \deg(Q), you should divide first:

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

where S(x) is the quotient polynomial and R(x) is the remainder polynomial with smaller degree than Q(x).

Why this matters: polynomials integrate easily, and the leftover fraction often becomes simpler (sometimes even a substitution-to-log pattern).

Example 1: long division leads to a log

Evaluate:

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

Step 1: Long divide x^2+1 by x+1.

• x^2/(x)=x, so subtract x(x+1)=x^2+x from x^2+0x+1 to get -x+1.
• (-x)/(x)=-1, so subtract -1(x+1)=-x-1 from -x+1 to get remainder 2.

So:

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

Step 2: 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

For the last term, substitution with u=x+1 gives a log:

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

Combine:

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

Common mistake: trying substitution immediately on the original fraction without dividing, which hides the simple polynomial part.

Example 2: long division simplifies the integrand structure

Evaluate:

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

Step 1: Divide 2x^3-3x+1 by x^2+1.

• Leading term: 2x^3/(x^2)=2x. Multiply back: 2x(x^2+1)=2x^3+2x. Subtract from 2x^3+0x^2-3x+1 to get -5x+1.

So:

\frac{2x^3-3x+1}{x^2+1} = 2x + \frac{-5x+1}{x^2+1}

Step 2: Split the remaining fraction:

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

Now each piece matches a basic pattern:

• \int 2x\,dx = x^2
• \int \frac{-5x}{x^2+1}dx is substitution with u=x^2+1, du=2x\,dx
• \int \frac{1}{x^2+1}dx is an arctangent pattern

Compute them:

\int \frac{-5x}{x^2+1}dx = -\frac{5}{2}\int \frac{2x}{x^2+1}dx = -\frac{5}{2}\ln|x^2+1|

And:

\int \frac{1}{x^2+1}dx = \arctan(x)

So the full antiderivative is:

x^2 - \frac{5}{2}\ln|x^2+1| + \arctan(x) + C

Two subtle points students miss:

• The remainder step is not optional—without it, you may never see the clean \ln|x^2+1| and \arctan(x) forms.
• \int \frac{1}{x^2+1}dx is not \ln|x^2+1|. That log appears when the numerator is proportional to 2x.

Completing the square to match standard integrals

Completing the square rewrites a quadratic into a shifted-square form:

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

This matters in integration because many important “recognition” integrals involve expressions like:

\int \frac{1}{x^2+a^2}dx

and the shifted version:

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

These connect to the inverse tangent function.

A standard result you should be comfortable using (and deriving via substitution) is:

\int \frac{1}{a^2+x^2}dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C \quad (a>0)

More generally:

\int \frac{1}{(x-h)^2+a^2}dx = \frac{1}{a}\arctan\left(\frac{x-h}{a}\right) + C \quad (a>0)

The key is to rewrite your quadratic denominator into that shape.

Example 3: completing the square leads to arctan

Evaluate:

\int \frac{1}{x^2+4x+8}\,dx

Step 1: Complete the square in the denominator.

x^2+4x+8 = (x^2+4x+4)+4 = (x+2)^2+4

So the integral becomes:

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

Step 2: Match the arctan form. Here 4=2^2, so a=2.

Use the shifted formula:

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

Common mistake: stopping at \arctan(x+2) without accounting for the scaling by 2.

Example 4: completing the square plus a constant factor

Evaluate:

\int \frac{3}{x^2-6x+13}\,dx

Step 1: Complete the square:

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

So:

\int \frac{3}{(x-3)^2+4}\,dx = 3\int \frac{1}{(x-3)^2+2^2}\,dx

Step 2: Use the formula:

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

A common error is to forget that the coefficient in front of the integral stays outside; another is to rewrite +4 as +(4)^2 instead of +(2)^2.

Choosing between long division and completing the square

Sometimes both appear in a single problem. A good decision process is:

• If you see a rational function with a “top-heavy” numerator, divide first.
• If you end up with a quadratic expression that doesn’t factor nicely (especially in a denominator), consider completing the square to reach a log or arctan form.

These techniques often prepare the integrand so that substitution (from the previous section) becomes obvious.

Exam Focus

• Typical question patterns:
  • “Evaluate an indefinite integral of a rational function” where you must perform long division before integrating.
  • “Integrate a function with a quadratic denominator” that doesn’t factor, pushing you to complete the square and use an arctan form.
  • Mixed problems where division creates terms like \frac{2x}{x^2+1} (log via substitution) and \frac{1}{x^2+1} (arctan).
• Common mistakes:
  • Skipping long division when \deg(P) \ge \deg(Q), which hides easy polynomial pieces.
  • Confusing the log pattern \int \frac{f'(x)}{f(x)}dx = \ln|f(x)|+C with the arctan pattern \int \frac{1}{x^2+a^2}dx.
  • Completing the square correctly but forgetting the scaling factor needed to match a^2 (leading to missing constants in the final answer).