The Power Rule For Antiderivatives

Introduction

An antiderivative reverses differentiation: a function $F$ is an antiderivative of $f$ on an interval if $F'(x)=f(x)$ throughout that interval. The earlier material on antiderivatives established this idea and introduced indefinite integral notation. Here we develop a useful rule for a basic family of functions: real powers of $x$.

The power rule is easy to apply once its two changes are remembered: increase the exponent by one, then divide by the new exponent. The condition on the original exponent matters. In particular, the rule in this form does not apply when the original exponent is negative one.

The Power Rule for Antiderivatives

For a real exponent $r$, the expression $x^r$ is a real-valued function for every $x>0$. This positive domain is sufficient for a rule that works uniformly for all real $r$. The interval of integration can be any interval contained in $(0,\infty)$.

Power rule for antiderivatives: If $r$ is a real number with $r\ne -1$, then on any interval in $(0,\infty)$, $$ \int x^r\,dx=\frac{x^{r+1}}{r+1}+C, $$ where $C$ is an arbitrary constant.

The exponent $r$ is the exponent in the integrand. The exponent in the antiderivative is $r+1$, and the denominator is that same new exponent. The constant $C$ is included because adding a constant does not change a derivative.

In this lesson, the rule is applied to a single power $x^r$. It is not yet a rule for integrating sums or constant multiples; those require separate properties of indefinite integration. Also, the symbol $\int x^r\,dx$ denotes a family of antiderivatives, not a definite integral with bounds.

Why the Exponent Increases and the Denominator Appears

The reason for the rule comes from the derivative power rule. For $x>0$, the derivative of $x^s$, where $s$ is real, is $s x^{s-1}$. To reverse this, we want an expression whose exponent becomes $r$ after differentiation. Raising $r$ by one gives that exponent, since $(r+1)-1=r$.

There is still a coefficient to choose. Suppose the candidate has the form $A x^{r+1}$, where $A$ is a constant. Differentiating gives $A(r+1)x^r$. To make this equal to $x^r$, the coefficient must satisfy $A(r+1)=1$. If $r\ne -1$, this equation has the solution $A=1/(r+1)$. This determines the formula.

\frac{d}{dx}\left(Ax^{r+1}\right) =A(r+1)x^r, \qquad A(r+1)=1 \quad\Longrightarrow\quad A=\frac{1}{r+1}.

This argument also explains the restriction: when $r=-1$, the coefficient equation would require $A\cdot 0=1$, which has no solution. The displayed power-rule formula therefore cannot be used for that exponent. The reciprocal function and its antiderivative require a separate treatment.

A Reliable Application Process

For an integrand that is exactly $x^r$, use the following sequence. Keeping the exponent change and coefficient together helps prevent errors.

Identify the exponent.
Write down the value of $r$ in $x^r$, and check that $r\ne -1$.

Increase it by one.
Use $r+1$ as the exponent in the antiderivative.

Divide by the new exponent.
Place $r+1$ in the denominator. The numerator and denominator must use the same new exponent.

Include the constant.
Write $+C$ to represent all antiderivatives on the interval.

Check by differentiation.
Differentiate the proposed antiderivative and confirm that the result is $x^r$.

Worked Examples

Worked Example: A Positive Integer Exponent

Find an antiderivative of $x^6$ on $(0,\infty)$.

The exponent is $r=6$, so $r+1=7$. The exponent is not $-1$, and the power rule gives

\int x^6\,dx =\frac{x^{6+1}}{6+1}+C =\frac{x^7}{7}+C.

Check the result by differentiating. Since $C$ is constant, its derivative is zero:

\frac{d}{dx}\left(\frac{x^7}{7}+C\right) =\frac{1}{7}\cdot 7x^6+0 =x^6.

The derivative is the original integrand, so the antiderivative checks.

Worked Example: An Irrational Exponent

Find an antiderivative of $x^{\sqrt{3}}$ on $(0,\infty)$.

Here $r=\sqrt{3}$. The rule applies because $\sqrt{3}\ne -1$. Increase the exponent by one and divide by that new exponent:

\int x^{\sqrt{3}}\,dx =\frac{x^{\sqrt{3}+1}}{\sqrt{3}+1}+C.

The exact form is preferable to a decimal approximation: there is no need to round the irrational exponent or coefficient. To check it, differentiate the proposed expression:

\frac{d}{dx}\left(\frac{x^{\sqrt{3}+1}}{\sqrt{3}+1}+C\right) =\frac{\sqrt{3}+1}{\sqrt{3}+1}x^{(\sqrt{3}+1)-1} =x^{\sqrt{3}}.

The coefficient simplifies to $1$, and the exponent simplifies to $\sqrt{3}$, recovering the integrand.

Worked Example: A Negative Real Exponent Other Than Negative One

Find an antiderivative of $x^{-\sqrt{2}}$ on $(0,\infty)$.

In this case $r=-\sqrt{2}$. The rule is allowed because $-\sqrt{2}\ne -1$. Add one to the exponent:

-\sqrt{2}+1=1-\sqrt{2}.

Now use $1-\sqrt{2}$ both as the new exponent and as the denominator:

\int x^{-\sqrt{2}}\,dx =\frac{x^{1-\sqrt{2}}}{1-\sqrt{2}}+C.

For the check, differentiate and simplify each factor:

\frac{d}{dx}\left(\frac{x^{1-\sqrt{2}}}{1-\sqrt{2}}+C\right) =\frac{1-\sqrt{2}}{1-\sqrt{2}}x^{(1-\sqrt{2})-1} =x^{-\sqrt{2}}.

Thus the formula works for this negative real exponent as well. The function and its antiderivative are being considered for $x>0$, where these real powers are defined.

How the Restriction and Domain Affect the Rule

There are two separate conditions to keep in view. First, the exponent must not equal $-1$, because the denominator $r+1$ would then be zero. Second, when $r$ is an arbitrary real number, the expression $x^r$ is taken here for $x>0$. So the rule gives an antiderivative on positive intervals, rather than claiming that every real power is defined for every real input.

The interval matters because an antiderivative is defined by a derivative relation throughout a specified interval. If the integrand is being considered on a smaller positive interval, the same formula applies there. The constant $C$ can take any real value; choosing different values produces different antiderivatives with the same derivative.

The exceptional exponent should not be forced into the formula. Substituting $r=-1$ would produce a denominator of zero, so it is not a valid application. The power rule as stated makes no claim for that case; a different rule is needed for the reciprocal function.

A Common Mistake: Dividing by the Old Exponent

A frequent error is to divide by $r$ without first increasing the exponent. For example, a mistaken calculation for $x^4$ might be

\int x^4\,dx\;\overset{\text{wrong}}{=}\;\frac{x^4}{4}+C.

This does not pass the derivative check:

\frac{d}{dx}\left(\frac{x^4}{4}+C\right) =\frac{4x^3}{4} =x^3,

and $x^3$ is not the original integrand $x^4$. The correction is to increase the exponent to $5$, then divide by $5$:

\int x^4\,dx=\frac{x^5}{5}+C, \qquad \frac{d}{dx}\left(\frac{x^5}{5}+C\right) =\frac{5x^4}{5}=x^4.

A useful check is to ask whether differentiating the proposed expression both reduces its exponent to the original one and cancels its denominator with the derivative's coefficient.

Why the Rule Is Useful

Real powers occur in many mathematical models, including functions that describe changing quantities and rates. The power rule provides a direct way to reverse differentiation for each single power that satisfies its condition. It also gives a clear test for a proposed answer: differentiate it and compare the result with the integrand.

For now, the important skill is using the rule accurately on one power at a time. Keep the domain in view, exclude $r=-1$, increase the exponent by one, divide by that new exponent, and include $C$. These checks make the method dependable without relying on a memorized pattern alone.

Check Your Understanding

Answer each question using the power rule and, where applicable, verify the result by differentiation.

Find an antiderivative of $x^9$ on $(0,\infty)$.
For $r=\sqrt{5}$, what exponent and denominator appear in the power-rule antiderivative of $x^r$?
Find an antiderivative of $x^{-\sqrt{3}}$ on $(0,\infty)$.
Why can the formula $\frac{x^{r+1}}{r+1}+C$ not be used when $r=-1$?
What is wrong with the claim $\int x^3\,dx=\frac{x^3}{3}+C$? Show the correction by differentiating.

The Power Rule For Antiderivatives

What You'll Learn

Introduction