Dove's LibraryDove's Library

6.8 - Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation

General vs. Particular Solutions

Multiple Functions, Same Derivative

Look at these three functions:
f(x)=x2+1g(x)=x2−5h(x)=x2+1000f(x) = x^2 + 1 \qquad g(x) = x^2 - 5 \qquad h(x) = x^2 + 1000
Take the derivative of each:
f′(x)=2xg′(x)=2xh′(x)=2xf'(x) = 2x \qquad g'(x) = 2x \qquad h'(x) = 2x
All three have the same derivative, which means the antiderivative of 2x2x has infinitely many possible answers.

Definitions

General solution: includes +C+C to represent all possible antiderivatives.
∫2x dx=x2+C\int 2x \, dx = x^2 + C
where CC is an arbitrary constant (the constant of integration).
Particular solution: one specific antiderivative with a determined value of CC.
∫2x dx=x2+1(when C=1)\int 2x \, dx = x^2 + 1 \quad \text{(when } C = 1\text{)}

Indefinite vs. Definite Integrals

Type
Notation
Result
Constant?
Indefinite
∫f(x) dx\int f(x) \, dx
General solution (a function)
Include +C+C
Definite
∫abf(x) dx\int_a^b f(x) \, dx
A number (area)
No +C+C, it cancels
Key difference: indefinite integrals represent a family of functions, while definite integrals give a specific numerical value.

Exponential Functions

Basic Exponential: exe^x

∫ex dx=ex+C\boxed{\int e^x \, dx = e^x + C}
Why? Because the derivative of exe^x is itself: ddx[ex]=ex\frac{d}{dx}[e^x] = e^x.

General Exponential: axa^x

Derivative (review):
ddx[ax]=axln⁡(a)\frac{d}{dx}[a^x] = a^x \ln(a)
Antiderivative:
∫ax dx=axln⁡(a)+C(a>0,a≠1)\boxed{\int a^x \, dx = \frac{a^x}{\ln(a)} + C} \quad (a > 0, a \neq 1)

Example 1

∫7x dx=7xln⁡(7)+C\int 7^x \, dx = \frac{7^x}{\ln(7)} + C

Example 2: Particular Solution

Find f(x)f(x) given f′(x)=ex−2x2f'(x) = e^x - 2x^2 and f(0)=4f(0) = 4.
Step 1: Find the general solution.
f(x)=∫(ex−2x2) dx=ex−2x33+Cf(x) = \int (e^x - 2x^2) \, dx = e^x - \frac{2x^3}{3} + C
Step 2: Use the initial condition to find CC. Plug in (0,4)(0, 4):
4=e0−2(0)33+C4 = e^0 - \frac{2(0)^3}{3} + C4=1−0+C4 = 1 - 0 + CC=3C = 3
Step 3: Write the particular solution.
f(x)=ex−2x33+3f(x) = e^x - \frac{2x^3}{3} + 3

Logarithmic Functions

The Natural Logarithm

∫1x dx=ln⁡∣x∣+C\boxed{\int \frac{1}{x} \, dx = \ln|x| + C}
Important: you have to include the absolute value. It’s what lets the domain include negative xx-values.
Why? Because ddx[ln⁡∣x∣]=1x\frac{d}{dx}[\ln|x|] = \frac{1}{x} for all x≠0x \neq 0.

When to Use Logarithms

Use the log form when you have ∫1x dx\int \frac{1}{x} \, dx or ∫x−1 dx\int x^{-1} \, dx.
Why the power rule fails here:
∫x−1 dx=x−1+1−1+1=x00=10which is undefined\int x^{-1} \, dx = \frac{x^{-1+1}}{-1+1} = \frac{x^0}{0} = \frac{1}{0} \quad \text{which is undefined}

Common Mistake

Wrong: ∫1x2 dx=ln⁡∣x2∣+C\int \frac{1}{x^2} \, dx = \ln|x^2| + C
Correct: ∫1x2 dx=∫x−2 dx=x−1−1+C=−1x+C\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C
Rule: only use ln⁡∣x∣\ln|x| when the exponent is exactly −1-1, which is the same as having 1x\frac{1}{x}.

Example 1: General Solution

∫(1−5x)dx=x−5ln⁡∣x∣+C\int \left(1 - \frac{5}{x}\right) dx = x - 5\ln|x| + C

Example 2: Particular Solution

Find f(x)f(x) given f(x)=∫(2x+6x−2)dxf(x) = \int \left(\dfrac{2}{x} + 6x^{-2}\right) dx and f(1)=−7f(1) = -7.
Step 1: Find the general solution.
f(x)=2ln⁡∣x∣−6x+Cf(x) = 2\ln|x| - \frac{6}{x} + C
Step 2: Use the initial condition.
−7=2ln⁡∣1∣−61+C-7 = 2\ln|1| - \frac{6}{1} + C−7=0−6+C-7 = 0 - 6 + CC=−1C = -1
Step 3: Write the particular solution.
f(x)=2ln⁡∣x∣−6x−1f(x) = 2\ln|x| - \frac{6}{x} - 1

Trigonometric Functions

Basic Trig Integrals

∫cos⁡x dx=sin⁡x+C\boxed{\int \cos x \, dx = \sin x + C}∫sin⁡x dx=−cos⁡x+C\boxed{\int \sin x \, dx = -\cos x + C}
Memory aid: work backwards from the derivatives.

Extended Trig Integrals

∫sec⁡2x dx=tan⁡x+C\boxed{\int \sec^2 x \, dx = \tan x + C}∫csc⁡2x dx=−cot⁡x+C\boxed{\int \csc^2 x \, dx = -\cot x + C}∫csc⁡xcot⁡x dx=−csc⁡x+C\boxed{\int \csc x \cot x \, dx = -\csc x + C}∫sec⁡xtan⁡x dx=sec⁡x+C\boxed{\int \sec x \tan x \, dx = \sec x + C}

Sign Pattern Rule

For derivatives: if the function starts with “c” (cosine, cotangent, cosecant), the derivative has a negative sign in front.
For integrals: if the answer starts with “c”, it has a negative sign in front.

Example: Particular Solution

Find yy given dydx=1−sin⁡x\frac{dy}{dx} = 1 - \sin x and y(π)=π−2y(\pi) = \pi - 2.
Step 1: Integrate.
y=∫(1−sin⁡x) dx=x+cos⁡x+Cy = \int (1 - \sin x) \, dx = x + \cos x + C
Step 2: Apply the condition.
π−2=π+cos⁡(π)+C\pi - 2 = \pi + \cos(\pi) + Cπ−2=π+(−1)+C\pi - 2 = \pi + (-1) + CC=−1C = -1
Step 3: Write the solution.
y=x+cos⁡x−1y = x + \cos x - 1

Inverse Trigonometric Functions

Pattern Recognition

Memory device: functions starting with “s” use square roots and subtraction.
∫11−x2 dx=sin⁡−1(x)+C=arcsin⁡(x)+C\boxed{\int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1}(x) + C = \arcsin(x) + C}∫1∣x∣x2−1 dx=sec⁡−1(x)+C=arcsec(x)+C\boxed{\int \frac{1}{|x|\sqrt{x^2-1}} \, dx = \sec^{-1}(x) + C = \text{arcsec}(x) + C}
Note: in sin⁡−1\sin^{-1}, the “i” looks like a “1,” which reminds you that the 1 comes first: 1−x2\sqrt{1-x^2}.
∫11+x2 dx=tan⁡−1(x)+C=arctan⁡(x)+C\boxed{\int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C = \arctan(x) + C}
For tangent: no “s,” so no square root, and you use addition instead of subtraction.

Summary Table

Integral Form
Antiderivative
11−x2\dfrac{1}{\sqrt{1-x^2}}
sin⁡−1(x)+C\sin^{-1}(x) + C
1∣x∣x2−1\dfrac{1}{\lvert x \rvert\sqrt{x^2-1}}
sec⁡−1(x)+C\sec^{-1}(x) + C
11+x2\dfrac{1}{1+x^2}
tan⁡−1(x)+C\tan^{-1}(x) + C

Algebraic Strategies

Strategy 1: Separate Rational Functions

Divide each term in the numerator by the denominator separately.
Example:
∫3x2+x−2x dx\int \frac{3x^2 + x - 2}{x} \, dx
Separate:
=∫(3x2x+xx−2x)dx=∫(3x+1−2x)dx= \int \left(\frac{3x^2}{x} + \frac{x}{x} - \frac{2}{x}\right) dx = \int \left(3x + 1 - \frac{2}{x}\right) dx
Integrate:
=3x22+x−2ln⁡∣x∣+C= \frac{3x^2}{2} + x - 2\ln|x| + C

Strategy 2: Distribute Before Integrating

When you’re multiplying expressions, distribute first to simplify.
Example:
∫3x(x+x2) dx\int 3x(\sqrt{x} + x^2) \, dx
Rewrite with exponents:
=∫3x(x1/2+x2) dx= \int 3x(x^{1/2} + x^2) \, dx
Distribute:
=∫(3x1⋅x1/2+3x1⋅x2) dx= \int (3x^{1} \cdot x^{1/2} + 3x^1 \cdot x^2) \, dx
Add the exponents:
=∫(3x3/2+3x3) dx= \int (3x^{3/2} + 3x^3) \, dx
Integrate:
=3⋅x5/25/2+3⋅x44+C=6x5/25+3x44+C= 3 \cdot \frac{x^{5/2}}{5/2} + 3 \cdot \frac{x^4}{4} + C = \frac{6x^{5/2}}{5} + \frac{3x^4}{4} + C

Functions Without a Closed-Form Antiderivative

Derivative rules let you differentiate any function built from powers, roots, trig, exponentials, and logs. Antiderivatives don’t work that way. Many functions have no closed-form antiderivative, meaning no formula built from those basic functions has them as its derivative. A few famous examples:
∫e−x2 dx∫sin⁡(x2) dx∫1+x3 dx\int e^{-x^2} \, dx \qquad \int \sin(x^2) \, dx \qquad \int \sqrt{1 + x^3} \, dx
These antiderivatives still exist (every continuous function has one). You just can’t write them as a formula, and no substitution or algebra trick will find one. When one shows up:
  • Definite integral: use your calculator. For example, ∫021+x3 dx≈3.241\int_0^2 \sqrt{1 + x^3} \, dx \approx 3.241.
  • Antiderivative as a function: write it as an accumulation function, like F(x)=∫0x1+t3 dtF(x) = \int_0^x \sqrt{1 + t^3} \, dt. By the Fundamental Theorem (Topic 6.4), F′(x)=1+x3F'(x) = \sqrt{1 + x^3}.

Key Takeaways

  • Indefinite integral: ∫f(x) dx\int f(x) \, dx gives a family of functions, with +C+C
  • General solution includes +C+C for the arbitrary constant
  • Particular solution has a specific value of CC found from an initial condition
  • Always include +C+C for indefinite integrals
  • Exponential: ∫ex dx=ex+C\int e^x \, dx = e^x + C and ∫ax dx=axln⁡a+C\int a^x \, dx = \frac{a^x}{\ln a} + C
  • Logarithm: ∫1x dx=ln⁡∣x∣+C\int \frac{1}{x} \, dx = \ln|x| + C, but only when the power is exactly −1-1
  • Basic trig: ∫cos⁡x dx=sin⁡x+C\int \cos x \, dx = \sin x + C and ∫sin⁡x dx=−cos⁡x+C\int \sin x \, dx = -\cos x + C
  • Sign rule: if the answer starts with “c,” it has a negative sign in front
  • Inverse trig: recognize the patterns with 1−x2\sqrt{1-x^2}, 1+x21+x^2, and so on
  • Absolute value is required in ln⁡∣x∣\ln|x| to extend the domain
  • Algebraic tricks: separate fractions or distribute before integrating
  • Finding CC: plug in the given point (x0,y0)(x_0, y_0) and solve
  • Check your work by taking the derivative of your answer
  • No formula? Functions like e−x2e^{-x^2} have antiderivatives you can’t write as a formula, so use a calculator or an accumulation function

AP® is a registered trademark of the College Board, which is not affiliated with Dove's Library and was not involved in the production of, and does not endorse, this website.

Providing free, high-quality AP notes to help students succeed, earn college credit, and reach their educational goals, ensuring everyone has access to the resources they need.

© 2026 Standard Learning Group, Inc., which operates Dove's Library. All rights reserved.