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Calculus I

Antiderivative Formulas

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Why This Matters

Antiderivatives are the foundation of integral calculus—every definite integral you'll evaluate, every area problem you'll solve, and every differential equation you'll tackle depends on your ability to reverse the differentiation process. You're being tested not just on memorizing formulas, but on recognizing which formula applies when you see a function and understanding why each formula works based on derivative rules you already know.

These formulas fall into distinct families: power functions, exponential functions, trigonometric functions, and inverse trigonometric functions. Each family follows predictable patterns rooted in differentiation. The key insight? Every antiderivative formula is just a derivative rule read backwards. Don't just memorize these formulas in isolation—know what differentiation rule each one reverses, and you'll be able to reconstruct any formula you forget on exam day.


Power Functions: The Foundation

The power rule for derivatives reverses cleanly into the power rule for antiderivatives. Since ddx[xn]=nxn1\frac{d}{dx}[x^n] = nx^{n-1}, we "undo" this by increasing the exponent and dividing.

Power Rule for Integration

  • xndx=1n+1xn+1+C\int x^n \, dx = \frac{1}{n+1} x^{n+1} + C where n1n \neq -1—this handles polynomials, roots (write as fractional exponents), and negative powers
  • The restriction n1n \neq -1 exists because plugging in n=1n = -1 creates division by zero; this case requires its own formula
  • Always add +C+ C to represent the constant of integration—the family of all functions whose derivative equals xnx^n

Natural Logarithm Rule

  • 1xdx=lnx+C\int \frac{1}{x} \, dx = \ln|x| + C—this fills the gap left by the power rule when n=1n = -1
  • The absolute value bars are essential because ln(x)\ln(x) is only defined for positive xx, but 1x\frac{1}{x} exists for all x0x \neq 0
  • Recognize disguised forms like x1dx\int x^{-1} \, dx—this is the same integral rewritten

Compare: Power Rule vs. Natural Log Rule—both handle expressions of the form xnx^n, but the log rule is the special case when n=1n = -1. If an FRQ gives you x1dx\int x^{-1} \, dx, don't mistakenly apply the power rule—that's a common error.


Exponential Functions: Self-Replicating Integrals

Exponential functions have the remarkable property that differentiation and integration preserve their form. The base determines the scaling factor.

Base ee Exponential

  • exdx=ex+C\int e^x \, dx = e^x + C—the exponential function exe^x is its own antiderivative, making it unique among all functions
  • This self-replicating property is why ee appears throughout differential equations modeling growth and decay
  • Watch for chain rule variantsekxdx=1kekx+C\int e^{kx} \, dx = \frac{1}{k}e^{kx} + C requires compensating for the inner derivative

General Exponential Base

  • axdx=axln(a)+C\int a^x \, dx = \frac{a^x}{\ln(a)} + C where a>0a > 0 and a1a \neq 1—the natural log of the base appears as a scaling factor
  • This formula reverses the derivative rule ddx[ax]=axln(a)\frac{d}{dx}[a^x] = a^x \ln(a), so we divide by ln(a)\ln(a) to compensate
  • When a=ea = e, note that ln(e)=1\ln(e) = 1, so this formula reduces to the simpler exe^x case

Compare: exdx\int e^x \, dx vs. axdx\int a^x \, dx—the exe^x formula is cleaner because ln(e)=1\ln(e) = 1 eliminates the denominator. On exams, convert to base ee when possible using ax=exln(a)a^x = e^{x \ln(a)}.


Basic Trigonometric Functions: Sine and Cosine Cycle

Trigonometric antiderivatives follow from the cyclic nature of trig derivatives. Since sine and cosine are derivatives of each other (with sign changes), their antiderivatives swap roles.

Sine Function

  • sin(x)dx=cos(x)+C\int \sin(x) \, dx = -\cos(x) + C—the negative sign appears because ddx[cos(x)]=sin(x)\frac{d}{dx}[\cos(x)] = -\sin(x)
  • Memory trick: integration of sine "brings down" cosine with a sign flip
  • Applications include oscillatory motion, wave equations, and any periodic phenomenon

Cosine Function

  • cos(x)dx=sin(x)+C\int \cos(x) \, dx = \sin(x) + C—no sign change here since ddx[sin(x)]=cos(x)\frac{d}{dx}[\sin(x)] = \cos(x)
  • The sine-cosine pair demonstrates the symmetry in trigonometric integration
  • Verify by differentiating: ddx[sin(x)]=cos(x)\frac{d}{dx}[\sin(x)] = \cos(x)

Secant Squared Function

  • sec2(x)dx=tan(x)+C\int \sec^2(x) \, dx = \tan(x) + C—this reverses the derivative ddx[tan(x)]=sec2(x)\frac{d}{dx}[\tan(x)] = \sec^2(x)
  • Frequently appears in problems involving trigonometric substitution and identities
  • Recognize equivalent forms like 1cos2(x)dx\int \frac{1}{\cos^2(x)} \, dx—same integral, different notation

Compare: sin(x)dx\int \sin(x) \, dx vs. cos(x)dx\int \cos(x) \, dx—both produce the other trig function, but only sine's antiderivative picks up a negative sign. Track signs carefully; this is a top source of errors on FRQs.


Advanced Trigonometric Functions: Logarithmic Results

Some trigonometric integrals produce logarithmic results rather than other trig functions. These arise from rewriting the integrand and applying substitution.

Tangent Function

  • tan(x)dx=lncos(x)+C\int \tan(x) \, dx = -\ln|\cos(x)| + C—equivalently written as lnsec(x)+C\ln|\sec(x)| + C
  • Derived by rewriting tan(x)=sin(x)cos(x)\tan(x) = \frac{\sin(x)}{\cos(x)} and using uu-substitution with u=cos(x)u = \cos(x)
  • The negative sign reflects that ddx[cos(x)]=sin(x)\frac{d}{dx}[\cos(x)] = -\sin(x)

Compare: tan(x)dx\int \tan(x) \, dx vs. sec2(x)dx\int \sec^2(x) \, dx—both involve tangent and secant, but one yields a logarithm while the other yields tangent directly. Know which is which; the squared secant is the cleaner case.


Inverse Trigonometric Functions: Recognizing the Patterns

These formulas produce inverse trig functions and arise from specific algebraic forms. The key is pattern recognition—spot the characteristic denominators.

Arctangent Pattern

  • 11+x2dx=arctan(x)+C\int \frac{1}{1+x^2} \, dx = \arctan(x) + C—this reverses ddx[arctan(x)]=11+x2\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}
  • The 1+x21 + x^2 denominator is your signal to use this formula
  • Generalizes to 1a2+x2dx=1aarctan(xa)+C\int \frac{1}{a^2+x^2} \, dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C

Arcsine Pattern

  • 11x2dx=arcsin(x)+C\int \frac{1}{\sqrt{1-x^2}} \, dx = \arcsin(x) + C—this reverses ddx[arcsin(x)]=11x2\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}
  • The 1x2\sqrt{1-x^2} denominator signals this formula; note the domain restriction x<1|x| < 1
  • Generalizes to 1a2x2dx=arcsin(xa)+C\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C

Compare: Arctangent vs. Arcsine patterns—both have "11" and "x2x^2" in the denominator, but arctangent has addition (1+x21 + x^2) while arcsine has subtraction under a square root (1x2\sqrt{1-x^2}). Memorize these signatures; they appear frequently on multiple choice.


Quick Reference Table

ConceptBest Examples
Power functionsxndx\int x^n \, dx, 1xdx\int \frac{1}{x} \, dx
Exponential functionsexdx\int e^x \, dx, axdx\int a^x \, dx
Basic trig (sine/cosine)sin(x)dx\int \sin(x) \, dx, cos(x)dx\int \cos(x) \, dx
Trig producing trigsec2(x)dx\int \sec^2(x) \, dx
Trig producing logarithmstan(x)dx\int \tan(x) \, dx
Inverse trig patterns11+x2dx\int \frac{1}{1+x^2} \, dx, 11x2dx\int \frac{1}{\sqrt{1-x^2}} \, dx
Self-replicating functionsexdx\int e^x \, dx
Special case (n=1n = -1)$$\int \frac{1}{x} , dx = \ln

Self-Check Questions

  1. Why does the power rule xndx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C fail when n=1n = -1, and which formula handles that case instead?

  2. Compare exdx\int e^x \, dx and 2xdx\int 2^x \, dx: what's the key difference in their antiderivatives, and why does one have a simpler form?

  3. If you see 14+x2dx\int \frac{1}{4 + x^2} \, dx on an exam, which antiderivative formula applies, and how would you adjust for the "4" instead of "1"?

  4. Both sin(x)dx\int \sin(x) \, dx and cos(x)dx\int \cos(x) \, dx produce the other function—which one picks up a negative sign, and how can you verify your answer?

  5. An FRQ asks you to evaluate 19x2dx\int \frac{1}{\sqrt{9-x^2}} \, dx. Identify the pattern, state the formula, and explain what adjustment the "9" requires.