Antiderivatives and indefinite integrals (basic rules) — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Definition of antiderivatives, indefinite integral notation, the constant of integration, and core basic antiderivative rules: power rule, constant multiple, sum/difference, trigonometric, and exponential rules for elementary functions.

You should already know: Derivative rules for all elementary functions, basic algebraic manipulation of polynomials and rational functions, standard function notation for derivatives.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Antiderivatives and indefinite integrals (basic rules)?

An antiderivative of a function $f(x)$ is a differentiable function $F(x)$ such that $F^{\prime }(x)=f(x)$ for all $x$ in the domain of $f$. The indefinite integral of $f(x)$ is the general form of all antiderivatives of $f$, written as $\int f(x)dx=F(x)+C$, where $C$ is the arbitrary constant of integration. This topic is the foundational starting point for all integration work in AP Calculus BC, per the College Board Course and Exam Description (CED), and makes up approximately 1-3% of the total exam score. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, most often as a building block for larger problems involving differential equations, area/volume calculations, or accumulation of change problems. Unlike definite integrals, which evaluate to a numerical quantity representing net change, indefinite integrals describe an entire family of functions that differ only by a vertical shift. Every advanced integration technique you will learn ultimately boils down to rewriting an integrand to fit one of these basic antiderivative rules, so mastery here is required for all subsequent integration topics.

2. Core Algebraic Antiderivative Rules

All basic antiderivative rules are derived by reversing corresponding derivative rules, starting with algebraic rules for polynomial and rational functions. The constant multiple rule for derivatives states $\frac{d}{dx}[kf(x)]=k{f}^{\prime }(x)$ for any constant $k$, so reversing this gives the integration equivalent: $$\int kf(x)dx=k\int f(x)dx$$ Similarly, the sum/difference rule for derivatives reverses to allow term-by-term integration: $$\int \left(f(x)\pm g(x)\right)dx=\int f(x)dx\pm \int g(x)dx$$ The most frequently used rule is the power rule for integration, which reverses the power rule for derivatives. Recall that $\frac{d}{dx}[x^n]=n{x}^{n-1}$, so solving for the antiderivative of $x^n$ gives: $$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C,n\ne -1$$ The restriction $n\ne -1$ is critical: if $n=-1$, $x^n=\frac{1}{x}$, and the formula would result in division by zero. For this special case, the antiderivative is $\ln |x|+C$, because the derivative of $\ln |x|$ is $\frac{1}{x}$ for all non-zero $x$.

Worked Example

Problem: Find the general indefinite integral $\int \left(4x^3-\frac{2}{x^2}+5\right)dx$.

1. First, rewrite all terms with exponents to fit the power rule: $\frac{2}{x^2}=2x^{-2}$, and $5=5x^0$, so the integrand becomes $4x^3-2x^{-2}+5x^0$.
2. Apply the sum/difference and constant multiple rules to split the integral into separate terms: $\int 4x^{3}dx-\int 2x^{-2}dx+\int 5x^{0}dx=4\int x^{3}dx-2\int x^{-2}dx+5\int x^{0}dx$.
3. Apply the power rule to each term:
   • $4\cdot \frac{x^{3+1}}{3+1}=4\cdot \frac{x^4}{4}=x^4$
   • $-2\cdot \frac{x^{-2+1}}{-2+1}=-2\cdot \frac{x^{-1}}{-1}=2x^{-1}=\frac{2}{x}$
   • $5\cdot \frac{x^{0+1}}{0+1}=5x$
4. Add the arbitrary constant of integration $C$ to get the general antiderivative.

Final result: $\int \left(4x^3-\frac{2}{x^2}+5\right)dx=x^4+\frac{2}{x}+5x+C$.

Exam tip: Always rewrite roots and rational terms as $x^n$ before applying the power rule to avoid mistakes with negative and fractional exponents.

3. Exponential and Logarithmic Antiderivative Rules

Next we reverse the derivative rules for exponential and logarithmic functions, which are used in every area of applied calculus from physics to economics. The simplest exponential rule comes from the fact that the derivative of $e^x$ is itself, so reversing gives: $$\int e^{x}dx=e^x+C$$ For exponential functions with a constant base $a\ne e$, recall that $\frac{d}{dx}[a^x]=a^x\ln a$, so we divide by $\ln a$ to reverse the derivative: $$\int a^{x}dx=\frac{a^x}{\ln a}+C,a>0,a\ne 1$$ We already encountered the antiderivative of the logarithmic base case $\frac{1}{x}$, but for the logarithm itself, the antiderivative is derived from the product rule for derivatives, resulting in: $$\int \ln xdx=x\ln x-x+C,x>0$$ A common point of confusion here is mixing up power functions (variable base, constant exponent: $x^n$) and exponential functions (constant base, variable exponent: $a^x$), which leads to misapplying the power rule to exponential functions and vice versa.

Worked Example

Problem: Evaluate the indefinite integral $\int \left(3e^x+2^x-\frac{4}{x}\right)dx$.

1. Split the integral by the sum/difference rule: $3\int e^{x}dx+\int 2^{x}dx-4\int \frac{1}{x}dx$.
2. Apply each rule: $\int e^{x}dx=e^x$, $\int 2^{x}dx=\frac{2^x}{\ln 2}$, $\int \frac{1}{x}dx=\ln |x|$.
3. Substitute back and multiply by constants: $3e^x+\frac{2^x}{\ln 2}-4\ln |x|$.
4. Add the constant of integration $C$.

Final result: $\int \left(3e^x+2^x-\frac{4}{x}\right)dx=3e^x+\frac{2^x}{\ln 2}-4\ln |x|+C$.

Exam tip: Always check if your function is a power function or exponential function before integrating; if the variable is in the exponent, use the exponential rule, not the power rule.

4. Trigonometric Antiderivative Rules

Every basic trigonometric derivative rule reverses directly to an antiderivative rule, and these are not provided on the AP Calculus formula sheet, so memorization is required. The core rules are all derived directly from reversing standard derivatives:

• $\frac{d}{dx}\sin x=\cos x⟹\int \cos xdx=\sin x+C$
• $\frac{d}{dx}(-\cos x)=\sin x⟹\int \sin xdx=-\cos x+C$
• $\frac{d}{dx}\tan x={\sec }^{2}x⟹\int {\sec }^{2}xdx=\tan x+C$
• $\frac{d}{dx}(-\cot x)={\csc }^{2}x⟹\int {\csc }^{2}xdx=-\cot x+C$
• $\frac{d}{dx}\sec x=\sec x\tan x⟹\int \sec x\tan xdx=\sec x+C$
• $\frac{d}{dx}(-\csc x)=\csc x\cot x⟹\int \csc x\cot xdx=-\csc x+C$

Negative signs are the most common source of error here, so always double-check your result by differentiating to confirm it matches the original integrand.

Worked Example

Problem: Find the general antiderivative of $f(x)=2\sin x-{\sec }^{2}x+3\sec x\tan x$.

1. Split into three separate integrals by the sum rule: $2\int \sin xdx-\int {\sec }^{2}xdx+3\int \sec x\tan xdx$.
2. Apply the trigonometric antiderivative rules to each term:
   • $\int \sin xdx=-\cos x$, so $2(-\cos x)=-2\cos x$
   • $\int {\sec }^{2}xdx=\tan x$, so $-\tan x$
   • $\int \sec x\tan xdx=\sec x$, so $3\sec x$
3. Combine terms and add the arbitrary constant $C$.

Final result: $\int (2\sin x-{\sec }^{2}x+3\sec x\tan x)dx=-2\cos x-\tan x+3\sec x+C$.

Exam tip: To avoid sign errors, quickly differentiate your final antiderivative to check that it matches the original integrand; this takes 10 seconds and catches most sign mistakes.

5. Common Pitfalls (and how to avoid them)

• Wrong move: $\int x^{-1}dx=\frac{x^0}{0}+C$, leaving it undefined or incorrectly writing it as $1+C$. Why: Students automatically apply the power rule without remembering the $n\ne -1$ exception. Correct move: Always check if $n=-1$ before applying the power rule, and use $\int \frac{1}{x}dx=\ln |x|+C$ for this case.
• Wrong move: $\int 3^{x}dx=\frac{3^{x+1}}{x+1}+C$, treating the base 3 as a power of $x$ and misapplying the power rule. Why: Confusion between exponential functions (variable exponent, constant base) and power functions (constant exponent, variable base). Correct move: Explicitly label functions as power ($x^n$) vs exponential ($a^x$) before integrating, and use the exponential antiderivative rule $\int a^{x}dx=\frac{a^x}{\ln a}+C$ for $a^x$.
• Wrong move: $\int 2\cos xdx=2\sin x$, omitting $+C$ when asked for the general antiderivative. Why: Students forget that indefinite integrals require the arbitrary constant, especially when carrying the result into a later part of a problem. Correct move: Add $+C$ to every indefinite integral result immediately after integrating, before moving to the next step of the problem.
• Wrong move: $\int \sin xdx=\cos x+C$, missing the negative sign. Why: Students memorize only the derivative rule and forget to reverse the sign correctly when integrating. Correct move: After integrating any trigonometric function, differentiate your result to confirm it matches the original integrand; for example, $\frac{d}{dx}[\cos x]=-\sin x\ne \sin x$, so you know a sign is wrong.
• Wrong move: $\int \frac{1}{x^3}dx=\frac{x^{-2}}{-2}+C$, then incorrectly rewriting this as $-2x^{-2}+C=-\frac{2}{x^2}+C$. Why: Students make an algebra mistake when simplifying the fraction after applying the power rule. Correct move: After calculating $\frac{x^{n+1}}{n+1}$, double-check the arithmetic with the denominator: for $n=-3$, $n+1=-2$, so $\frac{x^{-2}}{-2}=-\frac{1}{2x^2}$, not $-\frac{2}{x^2}$.
• Wrong move: $\int \frac{1}{x}dx=\ln x+C$, omitting the absolute value around $x$. Why: Students only work with positive $x$ in examples and forget that $\frac{1}{x}$ is defined for negative $x$, and $\ln x$ is not. Correct move: Always write the absolute value in the logarithm whenever you integrate $\frac{1}{x}$; it is required for full credit on the AP exam.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following is the general antiderivative of $f(x)=2\sqrt{x}-\frac{3}{\sqrt{x}}+5\cos x$? A) $\frac{4}{3}{x}^{3/2}-6\sqrt{x}+5\sin x+C$ B) $\frac{4}{3}{x}^{3/2}-\frac{3}{2}\sqrt{x}+5\sin x+C$ C) $2x^{3/2}-3x^{1/2}+5\sin x+C$ D) $\frac{4}{3}{x}^{3/2}-6\sqrt{x}-5\sin x+C$

Worked Solution: First, rewrite the integrand in power form to apply the power rule: $f(x)=2x^{1/2}-3x^{-1/2}+5\cos x$. Apply the power rule term-by-term: for $2x^{1/2}$, $n=1/2$, so the antiderivative is $2\cdot \frac{x^{3/2}}{3/2}=\frac{4}{3}{x}^{3/2}$. For $-3x^{-1/2}$, $n=-1/2$, so the antiderivative is $-3\cdot \frac{x^{1/2}}{1/2}=-6\sqrt{x}$. The antiderivative of $5\cos x$ is $5\sin x$, and we add the constant $C$. The final result matches option A. Correct answer: $\boxed{A}$.

Question 2 (Free Response)

Let $f^{\prime \prime }(x)=3x+\sin x-\frac{2}{x}$ for $x>0$. It is known that $f^{\prime }(1)=2$ and $f(1)=1$. (a) Find the general expression for $f^{\prime }(x)$. (b) Find the specific value of the constant of integration for $f^{\prime }(x)$, using the given initial condition. (c) Find the specific function $f(x)$ that satisfies all given conditions.

Worked Solution: (a) Integrate $f^{\prime \prime }(x)$ to get $f^{\prime }(x)$: $$f^{\prime }(x)=\int \left(3x+\sin x-\frac{2}{x}\right)dx=\frac{3}{2}{x}^2-\cos x-2\ln x+C_1$$ where $C_1$ is the arbitrary constant of integration, and the absolute value is dropped because $x>0$ is given.

(b) Substitute $f^{\prime }(1)=2$: $$2=\frac{3}{2}(1{)}^2-\cos (1)-2\ln (1)+C_1$$ Since $\ln (1)=0$, this simplifies to $2=\frac{3}{2}-\cos 1+C_1$, so $C_1=\frac{1}{2}+\cos 1$.

(c) Integrate $f^{\prime }(x)$ to get $f(x)$, using the antiderivative rule for $\ln x$: $$f(x)=\int \left(\frac{3}{2}{x}^2-\cos x-2\ln x+\frac{1}{2}+\cos 1\right)dx$$ $$=\frac{1}{2}{x}^3-\sin x-2(x\ln x-x)+\left(\frac{1}{2}+\cos 1\right)x+C_2$$ Simplify and substitute $f(1)=1$: $$1=\frac{1}{2}-\sin 1+2+\frac{1}{2}+\cos 1+C_2=3-\sin 1+\cos 1+C_2$$ $$C_2=-2+\sin 1-\cos 1$$ Final result: $$f(x)=\frac{1}{2}{x}^3-\sin x-2x\ln x+\left(\frac{5}{2}+\cos 1\right)x-2+\sin 1-\cos 1$$

Question 3 (Application / Real-World Style)

A small rocket is launched straight upward from the ground, with acceleration (in ${\text{m/s}}^2$) given by $a(t)=10+2t-\frac{1}{t+1}$ for $t\ge 0$, where $t$ is time in seconds after launch. At $t=0$, the velocity of the rocket is $0\text{ m/s}$, and it starts at position $y(0)=0\text{ m}$. Find the vertical position of the rocket 4 seconds after launch, rounded to one decimal place.

Worked Solution: Acceleration is the derivative of velocity, so integrate $a(t)$ to find $v(t)$: $$v(t)=\int \left(10+2t-\frac{1}{t+1}\right)dt=10t+t^2-\ln (t+1)+C$$ Apply $v(0)=0$: $0=0+0-\ln (1)+C⟹C=0$, so $v(t)=10t+t^2-\ln (t+1)$. Velocity is the derivative of position, so integrate to find $y(t)$: $$y(t)=\int \left(10t+t^2-\ln (t+1)\right)dt=5t^2+\frac{1}{3}{t}^3-(t+1)\ln (t+1)+(t+1)+C$$ Apply $y(0)=0$: $0=0+0-\ln (1)+1+C⟹C=-1$, so $y(t)=5t^2+\frac{1}{3}{t}^3-(t+1)\ln (t+1)+t$. Substitute $t=4$: $y(4)=5(16)+\frac{64}{3}-5\ln (5)+4\approx 80+21.333-8.047+4\approx 97.3$. Interpretation: Four seconds after launch, the rocket is approximately 97.3 meters above the ground.

7. Quick Reference Cheatsheet

8. What's Next

This chapter is the non-negotiable foundational prerequisite for every integration technique you will learn next in AP Calculus BC. The first topic that builds directly on these basic rules is u-substitution, which rewrites more complex integrands into a form that matches one of the basic rules you learned here. Without memorizing these basic antiderivative rules, you will not be able to recognize the simplified form after substitution, integration by parts, or partial fractions, making all subsequent integration work impossible. Beyond integration techniques, these rules are required to solve separable differential equations, calculate accumulation functions, areas, volumes, and arc lengths, all of which are heavily tested on the AP exam.

u-substitution for indefinite integrals separation of variables for differential equations integration by parts definite integrals and the Fundamental Theorem of Calculus