Antiderivatives & Indefinite Integrals

∫ Antiderivatives & Indefinite Integrals

Part 1 of 7 — What is an Antiderivative?

Definition

$\boxed{F'(x) = f(x) \iff \int f(x)\,dx = F(x) + C}$

An antiderivative of $f(x)$ is any function $F(x)$ whose derivative is $f(x)$.

The indefinite integral $\int f(x)\,dx$ represents the entire FAMILY of antiderivatives.

Key Concept: The "$+C$" is NOT optional! Since the derivative of any constant is 0, there are infinitely many antiderivatives. For example, $x^2$, $x^2 + 5$, and $x^2 - 3$ are ALL antiderivatives of $2x$.

Power Rule for Integration

$\boxed{\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)}$

Mnemonic: "Add one to the exponent, divide by the new exponent."

This reverses the Power Rule for differentiation. Quick check: $\frac{d}{dx}\left[\frac{x^{n+1}}{n+1}\right] = \frac{(n+1)x^n}{n+1} = x^n$ ✓

Worked Examples

AP Tip: Always CHECK your answer by differentiating! If $F'(x) = f(x)$, your integral is correct.

The Linearity Property

$\boxed{\int [af(x) + bg(x)]\,dx = a\int f(x)\,dx + b\int g(x)\,dx}$

Example: $\int (3x^2 - 5x + 4)\,dx = 3 \cdot \frac{x^3}{3} - 5 \cdot \frac{x^2}{2} + 4x + C = x^3 - \frac{5x^2}{2} + 4x + C$

Power Rule for Integration 🎯

Special Case: $n = -1$

The Power Rule breaks down when $n = -1$ (division by zero!):

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

Key Fact: Absolute value is REQUIRED. $\ln(x)$ is only defined for $x > 0$, but $\frac{1}{x}$ is defined for all $x \neq 0$. The absolute value makes the antiderivative valid for negative $x$ too.

Common Mistakes

Mixed Power Rule 🎯

Match each integral with its result. 🔍

Compute the coefficient. ✍️

Key Takeaways — Part 1

$\boxed{\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)}$

Up Next: Part 2 — Essential Antiderivative Formulas (trig, exponential, and more).

∫ Antiderivatives

Part 2 of 7 — Essential Antiderivative Formulas

The Complete Table

$\boxed{\text{Memorize these — they are the building blocks of ALL integration}}$

Key Fact: $e^x$ is its OWN antiderivative — the only function (up to constants) with this property!

Trigonometric Antiderivatives

Pattern: The "co-" functions (cosine, cosecant, cotangent) always get a NEGATIVE sign when integrating.

Inverse Trig Antiderivatives

AP Tip: These two inverse trig forms appear often on the AP exam. Know them cold!

Essential Antiderivatives 🎯

Linearity of Integration

$\boxed{\int [af(x) + bg(x)]\,dx = a\int f(x)\,dx + b\int g(x)\,dx}$

Worked Examples

Example 1: $\int (5\cos x - 3e^x + \frac{2}{x})\,dx$

$= 5\sin x - 3e^x + 2\ln|x| + C$

Example 2: $\int (\sec^2 x - \csc^2 x)\,dx$

$= \tan x - (-\cot x) + C = \tan x + \cot x + C$

What You CANNOT Do

Key Concept: Integration is LINEAR (constants and sums), but NOT multiplicative. You can only split sums and pull out constants.

Trig & Exponential Integrals 🎯

Match each integral to its result. 🔍

Evaluate the integral. ✍️

Key Takeaways — Part 2

$\boxed{\text{"Co-" functions get a negative: } \int \sin x = -\cos x, \quad \int \csc^2 x = -\cot x}$

Up Next: Part 3 — Initial Value Problems.

∫ Antiderivatives

Part 3 of 7 — Initial Value Problems (IVPs)

Finding a Specific Antiderivative

An initial condition pins down the exact value of $C$:

$\boxed{\text{General antiderivative} + \text{initial condition} = \text{particular solution}}$

Worked Example

Given: $f'(x) = 3x^2 - 4x + 1$ and $f(0) = 5$. Find $f(x)$.

Step 1: $f(x) = \int (3x^2 - 4x + 1)\,dx = x^3 - 2x^2 + x + C$

Step 2: $f(0) = 0 - 0 + 0 + C = 5$

Step 3: $C = 5$

$\boxed{f(x) = x^3 - 2x^2 + x + 5}$

Check: $f'(x) = 3x^2 - 4x + 1$ ✓ and $f(0) = 5$ ✓

Position-Velocity-Acceleration

$\boxed{a(t) \xrightarrow{\int} v(t) \xrightarrow{\int} s(t)}$

Each integration introduces a NEW constant, determined by initial conditions.

Worked Example: Free Fall

A ball is thrown upward at 64 ft/s from height 80 ft. Find $s(t)$.

$a(t) = -32$ $ft/s^{2}$ (gravity)

Step 1: $v(t) = \int (-32)\,dt = -32t + C_1$

• $v(0) = 64$ → $C_1 = 64$
• So $v(t) = -32t + 64$

Step 2: $s(t) = \int (-32t + 64)\,dt = -16t^2 + 64t + C_2$

• $s(0) = 80$ → $C_2 = 80$

$\boxed{s(t) = -16t^2 + 64t + 80}$

Key Fact: In free fall, $a = -32$ $ft/s^{2}$ or $a = -9.8$ $m/s^{2}$. The negative sign means downward. Two initial conditions are needed: $v(0)$ and $s(0)$.

Initial Value Problems 🎯

Multiple Initial Conditions

When given $f''(x)$, you need TWO initial conditions (one for each integration):

AP-Style Example

$f''(x) = 12x - 4$, $f'(1) = 3$, $f(2) = 10$. Find $f(x)$.

First integration: $f'(x) = 6x^2 - 4x + C_1$

• $f'(1) = 6 - 4 + C_1 = 3$ → $C_1 = 1$
• $f'(x) = 6x^2 - 4x + 1$

Second integration: $f(x) = 2x^3 - 2x^2 + x + C_2$

• $f(2) = 16 - 8 + 2 + C_2 = 10$ → $C_2 = 0$

$\boxed{f(x) = 2x^3 - 2x^2 + x}$

AP Tip: When an IVP asks for a SPECIFIC value like $f(3)$, you can sometimes use the definite integral: $f(3) = f(2) + \int_2^3 f'(x)\,dx$, which may be faster.

Motion IVPs 🎯

Solve the IVP step by step. 🔍

$f'(x) = 4x - 3$, $f(2) = 7$

Solve the IVP. ✍️

Key Takeaways — Part 3

$\boxed{f'(x) + \text{initial condition} \xrightarrow{\int} \text{unique } f(x)}$

Up Next: Part 4 — Algebraic Manipulation Before Integrating.

∫ Antiderivatives

Part 4 of 7 — Rewriting Before Integrating

The Strategy

$\boxed{\text{Can't integrate directly?} \to \text{Rewrite algebraically} \to \text{Then integrate}}$

Many integrals look hard but become easy after algebraic manipulation:

Key Concept: You CANNOT integrate products or quotients by integrating top and bottom separately. You must rewrite into a SUM first.

Technique 1: Expand Products

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

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

Technique 2: Split Fractions

$\int \frac{x^3 + 2x}{x^2}\,dx = \int \left(x + \frac{2}{x}\right)\,dx = \frac{x^2}{2} + 2\ln|x| + C$

$\int \frac{x^2 - 4x + 1}{\sqrt{x}}\,dx = \int (x^{3/2} - 4x^{1/2} + x^{-1/2})\,dx$ $= \frac{2x^{5/2}}{5} - \frac{8x^{3/2}}{3} + 2x^{1/2} + C$

Technique 3: Rewrite Radicals

$\int \frac{3}{\sqrt{x}}\,dx = \int 3x^{-1/2}\,dx = 6\sqrt{x} + C$

$\int \sqrt[3]{x^2}\,dx = \int x^{2/3}\,dx = \frac{x^{5/3}}{5/3} + C = \frac{3}{5}x^{5/3} + C$

AP Tip: Always convert to $x^n$ form before applying the Power Rule. Roots, reciprocals, and radicals are just fractional/negative exponents.

Simplify Then Integrate 🎯

Technique 4: Trig Identities

Sometimes you need a trig identity before integrating:

Key Fact: $\int \tan x\,dx$ and $\int \sec x\,dx$ require techniques you'll see later ($u$-substitution). For now, focus on recognizing when an identity simplifies the integrand.

Decision Flowchart

Trig & Advanced Rewriting 🎯

Identify the correct rewriting technique. 🔍

Evaluate the integral. ✍️

Key Takeaways — Part 4

$\boxed{\text{Rewrite} \to \text{Integrate term by term} \to \text{Simplify}}$

Up Next: Part 5 — Inverse Trig Antiderivatives.

∫ Antiderivatives

Part 5 of 7 — Inverse Trig Antiderivatives

The Two Essential Forms (AB Exam)

$\boxed{\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\left(\frac{x}{a}\right) + C}$

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

Key Fact: The arcsecant formula ($\int \frac{1}{x\sqrt{x^2-a^2}}$) is BC only. AB students need ONLY arcsin and arctan forms.

How to Recognize Them

Careful with signs: $\sqrt{x^2 - a^2}$ is NOT the arcsin form (notice the minus is flipped!).

Worked Examples

Example 1: $\int \frac{1}{\sqrt{9-x^2}}\,dx$

Here $a^2 = 9$, so $a = 3$: $\arcsin\left(\frac{x}{3}\right) + C$

Example 2: $\int \frac{1}{4 + x^2}\,dx$

Here $a^2 = 4$, so $a = 2$: $\frac{1}{2}\arctan\left(\frac{x}{2}\right) + C$

Example 3: $\int \frac{3}{\sqrt{1 - 4x^2}}\,dx$

Rewrite: $\frac{3}{\sqrt{1-(2x)^2}}$. Let $u = 2x$, $du = 2\,dx$:

$\frac{3}{2}\int \frac{du}{\sqrt{1-u^2}} = \frac{3}{2}\arcsin(u) + C = \frac{3}{2}\arcsin(2x) + C$

Completing the Square

Sometimes you need to complete the square first:

$\int \frac{1}{x^2 + 6x + 13}\,dx$

$x^2 + 6x + 13 = (x+3)^2 + 4$. Now it's arctan form!

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

AP Tip: If the denominator is a quadratic that doesn't factor, try completing the square to reveal an inverse trig form.

Inverse Trig Integrals 🎯

Don't Confuse These!

Key Concept: The inverse trig forms ONLY work when the numerator is a CONSTANT. If there's an $x$ in the numerator, it's a $u$-substitution problem instead!

Distinguish the Forms 🎯

Classify each integral. 🔍

Evaluate the definite integral. ✍️

Key Takeaways — Part 5

$\boxed{\frac{1}{\sqrt{a^2-x^2}} \to \arcsin\left(\frac{x}{a}\right) \qquad \frac{1}{a^2+x^2} \to \frac{1}{a}\arctan\left(\frac{x}{a}\right)}$

Up Next: Part 6 — Problem-Solving Workshop.

∫ Antiderivatives

Part 6 of 7 — Mixed Practice Workshop

The Real Challenge: Choosing the Right Tool

On the AP Exam, nobody tells you WHICH rule to use. You must:

1. Look at the integrand's structure
2. Classify it (power rule? trig? inverse trig? rewrite first?)
3. Apply the correct formula
4. Check by differentiating

Decision Flowchart

Key Concept: Most "hard" antiderivatives are actually easy formulas in disguise — you just need to rewrite first!

Mixed Worked Examples

Example 1: $\int \left(\frac{3}{x} + 4\cos x - x^5\right)dx$

Split: $3\ln|x| + 4\sin x - \frac{x^6}{6} + C$

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

Rewrite: $\int (x + 2x^{-1} - x^{-2})\,dx = \frac{x^2}{2} + 2\ln|x| + \frac{1}{x} + C$

Example 3: $\int (\sqrt{x} + e^x + \sec x \tan x)\,dx$

$= \frac{2}{3}x^{3/2} + e^x + \sec x + C$

Example 4 (IVP): $f'(x) = 6x^2 - 4x + 1$, $f(1) = 3$.

$f(x) = 2x^3 - 2x^2 + x + C$. $f(1) = 2 - 2 + 1 + C = 3 \Rightarrow C = 2$.

$f(x) = 2x^3 - 2x^2 + x + 2$

AP Tip: Always verify by differentiating. If $\frac{d}{dx}[\text{your answer}] = \text{integrand}$, you're correct!

Mixed Antiderivative Problems — Set 1 🎯

Mixed Antiderivative Problems — Set 2 🎯

Common AP Mistakes on Mixed Problems

Key Fact: The most common FRQ error is forgetting $+C$ on indefinite integrals. On the AP Exam, this can cost you a point!

Identify the technique and compute. 🔍

Solve the IVP. ✍️

Key Takeaways — Part 6

$\boxed{\text{Step 1: Classify} \to \text{Step 2: Apply} \to \text{Step 3: Verify by differentiating}}$

Up Next: Part 7 — Comprehensive Assessment.

∫ Antiderivatives — Comprehensive Review

Part 7 of 7 — Final Assessment

Complete Formula Reference

Quick-Reference Decision Guide

$\boxed{\text{Read} \to \text{Rewrite (if needed)} \to \text{Recognize the form} \to \text{Apply formula} \to \text{Add } +C}$

Top AP Exam Mistakes — Antiderivatives

Key Fact: On FRQs, you get a "linkage point" for correctly connecting your antiderivative to the initial condition. Show ALL steps: general solution → plug in IC → solve for $C$ → write particular solution.

Final Assessment — Set 1 🎯

Final Assessment — Set 2 🎯

AP-Style Mixed Classification 🔍

Final Challenge ✍️

Antiderivatives — Complete! ✅

You have mastered:

AP Exam Checklist

• ✅ Can I recognize ALL basic antiderivative forms?
• ✅ Can I rewrite integrands to match known forms?
• ✅ Do I always include $+C$ for indefinite integrals?
• ✅ Can I solve IVPs with one or two initial conditions?
• ✅ Can I distinguish inverse trig from $u$-sub cases?
• ✅ Do I verify answers by differentiating?

$\boxed{\text{Integration is the REVERSE of differentiation — always check by differentiating!}}$