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Calculus AB · Unit 6 · 6.8
AP Calculus · Integration and Accumulation of Change
Antiderivatives and Indefinite Integrals
Every derivative rule read backwards — plus a constant, and one genuinely exceptional exponent.
key idea$\int x^{n}\,dx=\frac{x^{\,n+1}}{n+1}+C\quad (n\neq-1)$
The rule
Antidifferentiation is differentiation run backwards, so every derivative rule you know supplies an integral rule. Two things are new: the $+C$, and the fact that $n=-1$ needs its own answer because $\tfrac{x^{0}}{0}$ is meaningless.
| $\displaystyle\int f(x)\,dx$ | Result | Check by differentiating |
|---|
| $x^{n}$, $n\neq-1$ | $\dfrac{x^{\,n+1}}{n+1}+C$ | $\dfrac{(n+1)x^{n}}{n+1}=x^{n}$ |
| $\dfrac1x$ | $\ln\lvert x\rvert+C$ | the exception to the power rule |
| $e^{x}$ | $e^{x}+C$ | unchanged, as always |
| $\cos x$ | $\sin x+C$ | $\left(\sin x\right)'=\cos x$ |
| $\sin x$ | $-\cos x+C$ | the minus moves to the answer |
| $\sec^{2}x$ | $\tan x+C$ | — |
| $\sec x\tan x$ | $\sec x+C$ | — |
| $\dfrac{1}{1+x^{2}}$ | $\arctan x+C$ | — |
| $\dfrac{1}{\sqrt{1-x^{2}}}$ | $\arcsin x+C$ | — |
The right-hand column is the only verification you ever need: differentiate your answer. If it returns the integrand, you are correct.
AntiderivativeAny $F$ with $F'=f$. There are infinitely many, differing by a constant.
Indefinite integral$\int f(x)\,dx$ — the whole family $F(x)+C$, not a number.
Definite integral$\int_a^b f(x)\,dx$ — a single number. The $+C$ cancels.
Particular antiderivativeThe one member of the family selected by an initial condition.
Pattern recognition
Rewrite firstroots and reciprocals become powers
Split freelylinearity works for antiderivatives too
Always verifydifferentiate the answer — it must return the integrand
Worked examples
EX 1
Rewrite, then integrate
$\int\left(\sqrt{x}+\dfrac{3}{x^{2}}\right)dx$Evaluate $\displaystyle\int\left(\sqrt{x}+\dfrac{3}{x^{2}}\right)dx$.
1
Rewrite as powers
$\sqrt{x}=x^{1/2}$ and $\dfrac{3}{x^{2}}=3x^{-2}$.
2
Integrate the first term
$\dfrac{x^{3/2}}{3/2}=\dfrac{2}{3}x^{3/2}$.
3
Integrate the second
$3\cdot\dfrac{x^{-1}}{-1}=-3x^{-1}$.
4
Assemble with one constant
$\dfrac{2}{3}x^{3/2}-\dfrac{3}{x}+C$ — a single $C$ covers the whole expression.
5
Verify
Differentiating gives $x^{1/2}+3x^{-2}$, the original integrand. ✓
EX 2
The exceptional exponent
$n=-1$Evaluate $\displaystyle\int\dfrac{5}{x}\,dx$ and explain why the power rule fails.
1
Try the power rule
$n=-1$ would give $\dfrac{x^{0}}{0}$ — division by zero, so the rule cannot apply.
2
Use the logarithm instead
$\displaystyle\int\dfrac{1}{x}\,dx=\ln\lvert x\rvert+C$.
3
Pull out the constant
$5\ln\lvert x\rvert+C$.
4
Justify the absolute value
$\ln x$ is undefined for $x\lt 0$, but $\tfrac1x$ is perfectly defined there. The absolute value extends the antiderivative to the whole domain.
5
Verify on the negative side
For $x\lt 0$: $\dfrac{d}{dx}\ln(-x)=\dfrac{-1}{-x}=\dfrac1x$. ✓
EX 3
Find the particular antiderivative
initial conditionFind $f$ given $f'(x)=6x^{2}-4$ and $f(1)=5$.
1
Antidifferentiate
$f(x)=2x^{3}-4x+C$.
2
Use the condition
$f(1)=2-4+C=5$.
3
Solve for $C$
$-2+C=5$, so $C=7$.
4
Write the answer
$f(x)=2x^{3}-4x+7$.
5
Why $C$ mattered here
In a definite integral the constant cancels. Here it does not — the condition selects one curve from the family.
EX 4
Simplify before integrating
$\int\dfrac{x^{3}+2x}{x}\,dx$Evaluate $\displaystyle\int\dfrac{x^{3}+2x}{x}\,dx$.
1
Resist looking for a rule
There is no quotient rule for integration. Check whether the fraction simplifies.
2
Divide through
$\dfrac{x^{3}+2x}{x}=x^{2}+2$ for $x\neq0$.
3
Integrate term by term
$\dfrac{x^{3}}{3}+2x+C$.
4
Verify
Differentiating gives $x^{2}+2$. ✓
5
The general lesson
Integration has no product or quotient rule. When you meet one, look for an algebraic simplification first.
Watch out
Never omit the $+C$
An indefinite integral is a family of functions. Dropping the constant is a standard deduction on free response.
$n=-1$ is not covered by the power rule
$\int x^{-1}dx=\ln\lvert x\rvert+C$. Applying the power rule there produces division by zero.
Keep the absolute value in $\ln\lvert x\rvert$
Without it the antiderivative is wrong for negative $x$, where the integrand is perfectly well defined.
There is no product or quotient rule for integrals
Simplify algebraically, or reach for substitution. Guessing a ‘reverse product rule’ produces nonsense.