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Antiderivatives and Substitution · Tutorial 484 of 999

Integrating Negative and Fractional Powers

Rewrite radicals and reciprocals as powers, integrate when the power rule applies, and check the domain before accepting an antiderivative.

Intermediate 9 min read

What You'll Learn

  • Rewrite radicals and reciprocals using rational and negative exponents
  • Apply the power rule to negative and fractional powers
  • Identify the domains of the original integrand and its antiderivative
  • Understand why the exponent negative one is excluded from the power rule
  • Verify an antiderivative by differentiating it

From Radicals and Reciprocals to Powers

Radicals and reciprocals can make an integrand look different from the powers used in the power rule for antiderivatives. Rewriting them as powers often reveals that the rule applies directly. The important checks are whether the rewritten expression is equivalent on the domain in question and whether its exponent is negative one.

Definition: For an exponent \(r\), an antiderivative of \(f\) on an interval \(I\) is a function \(F\) such that \(F'(x)=f(x)\) for every \(x\) in \(I\). The notation \(\int f(x)\,dx\) represents the family of antiderivatives on the interval being considered.

For positive \(x\), rational powers have their usual real meanings. In particular, \(x^{1/n}=\sqrt[n]{x}\), and \(x^{-a}=1/x^a\) when the denominator is nonzero. Thus, for example, \(\sqrt{x}=x^{1/2}\), \(1/\sqrt{x}=x^{-1/2}\), and \(1/x^3=x^{-3}\). These rewrites preserve the original expression only where the original expression is defined.

The Power Rule and Its Exception

The power rule for antiderivatives says that if \(r\ne -1\), then a power \(x^r\) has an antiderivative obtained by increasing the exponent by one and dividing by the new exponent. For positive \(x\), this rule applies to integer and rational exponents, including negative and fractional ones.

Power rule for antiderivatives: If \(r\ne -1\), then
$$ \int x^r\,dx=\frac{x^{r+1}}{r+1}+C, $$
on an interval where \(x^r\) is defined and differentiable. Here \(C\) is an arbitrary constant.

The condition \(r\ne -1\) is essential. When \(r=-1\), the proposed denominator \(r+1\) would be zero, so the formula does not produce an antiderivative. The integral of \(x^{-1}=1/x\) requires a distinct logarithmic antiderivative rule; the power rule alone does not evaluate it.

Why the Rule Works

The power rule for antiderivatives reverses the derivative power rule. On an interval where the relevant powers are defined and differentiable, the derivative of \(x^{r+1}\) is \((r+1)x^r\). If \(r\ne -1\), dividing by the nonzero constant \(r+1\) cancels that factor.

1
Choose a candidate.
For \(r\ne -1\), set \(F(x)=x^{r+1}/(r+1)\).
2
Differentiate it.
The derivative power rule gives \(F'(x)=\frac{1}{r+1}(r+1)x^r\).
3
Simplify and verify.
Since \(r+1\ne 0\), \(F'(x)=x^r\), so \(F\) is an antiderivative of \(x^r\).

This verifies the formula wherever the power and its derivative are defined. Adding a constant does not change the derivative, because the derivative of a constant is zero. On a connected interval, all antiderivatives of the same function differ by a constant.

Worked Example: Integrating a Square Root

Worked Example: Integrating a Square Root

Find \(\int \sqrt{x}\,dx\) for \(x>0\), then check the result.

First rewrite the radical as a power:

$$ \sqrt{x}=x^{1/2}. $$

The exponent is \(r=\frac12\), which is not \(-1\). Increase the exponent by one and divide by the result:

$$ \begin{aligned} \int \sqrt{x}\,dx &=\int x^{1/2}\,dx\\ &=\frac{x^{1/2+1}}{1/2+1}+C\\ &=\frac{x^{3/2}}{3/2}+C\\ &=\frac{2}{3}x^{3/2}+C. \end{aligned} $$

To check, differentiate the proposed antiderivative:

$$ \frac{d}{dx}\left(\frac{2}{3}x^{3/2}+C\right) =\frac{2}{3}\cdot\frac{3}{2}x^{1/2} =x^{1/2} =\sqrt{x}. $$

The derivative matches the original integrand, so the result is correct on \(x>0\).

Worked Example: Integrating a Reciprocal Radical

Worked Example: Integrating a Reciprocal Radical

Find \(\int \frac{1}{\sqrt{x}}\,dx\) for \(x>0\), and verify the answer.

The reciprocal becomes a negative exponent:

$$ \frac{1}{\sqrt{x}}=\frac{1}{x^{1/2}}=x^{-1/2}. $$

Here \(r=-\frac12\), so \(r+1=\frac12\ne 0\). Apply the rule:

$$ \begin{aligned} \int \frac{1}{\sqrt{x}}\,dx &=\int x^{-1/2}\,dx\\ &=\frac{x^{-1/2+1}}{-1/2+1}+C\\ &=\frac{x^{1/2}}{1/2}+C\\ &=2x^{1/2}+C\\ &=2\sqrt{x}+C. \end{aligned} $$

Differentiate to check:

$$ \frac{d}{dx}\left(2\sqrt{x}+C\right) =2\cdot\frac12 x^{-1/2} =x^{-1/2} =\frac{1}{\sqrt{x}}. $$

The original integrand is undefined at \(x=0\), so the result applies on the interval \(x>0\), not at zero.

Negative Fractional Powers and Domain

A negative fractional power combines two features: the exponent is fractional, and the expression involves a reciprocal. For example, \(x^{-1/2}=1/\sqrt{x}\) is defined for \(x>0\). The square root requires \(x\ge 0\), while the reciprocal excludes \(x=0\). Both restrictions matter, leaving \(x>0\).

Not every fractional power requires a positive base. When a rational exponent is written in lowest terms with an odd denominator, its real value can also be defined for negative \(x\). For example, \(x^{1/3}=\sqrt[3]{x}\) is real for every real \(x\). If the exponent has an even denominator, the corresponding real root requires a nonnegative base. A negative exponent additionally excludes zero because it places the power in a denominator.

Worked Example: Integrating a Cube-Root Power

Find \(\int x^{1/3}\,dx\) for real \(x\), and check the result.

The exponent \(1/3\) has an odd denominator, so \(x^{1/3}\) is real for every real \(x\). The power rule gives:

$$ \begin{aligned} \int x^{1/3}\,dx &=\frac{x^{1/3+1}}{1/3+1}+C\\ &=\frac{x^{4/3}}{4/3}+C\\ &=\frac34x^{4/3}+C. \end{aligned} $$

Differentiate the result:

$$ \frac{d}{dx}\left(\frac34x^{4/3}+C\right) =\frac34\cdot\frac43x^{1/3} =x^{1/3}. $$

This confirms the antiderivative wherever the derivative rule applies. At \(x=0\), the proposed function is also differentiable: its difference quotient there is \(\frac34h^{1/3}\), which tends to \(0\) as \(h\) tends to \(0\). The integrand has value \(0\) there, so the derivative agrees at zero as well.

Intervals and Constants of Integration

Domain restrictions can split the real line into separate intervals. For instance, \(x^{-3}=1/x^3\) is undefined at zero. Its domain is \((-\infty,0)\cup(0,\infty)\), so an antiderivative must be considered separately on those two intervals. The constants on the intervals need not be the same.

Worked Example: Integrating a Negative Integer Power

Find antiderivatives of \(x^{-3}\) on both intervals in its domain.

The exponent is \(-3\), not \(-1\), so the power rule applies. Increasing the exponent by one gives \(-2\):

$$ \begin{aligned} \int x^{-3}\,dx &=\frac{x^{-3+1}}{-3+1}+C\\ &=\frac{x^{-2}}{-2}+C\\ &=-\frac12x^{-2}+C\\ &=-\frac{1}{2x^2}+C. \end{aligned} $$

Check by differentiation:

$$ \frac{d}{dx}\left(-\frac12x^{-2}+C\right) =-\frac12(-2)x^{-3} =x^{-3} =\frac{1}{x^3}. $$

The formula is valid for \(x\ne0\). More precisely, the full family on the domain can use one constant on each interval: \(F(x)=-1/(2x^2)+C_-\) for \(x<0\), and \(F(x)=-1/(2x^2)+C_+\) for \(x>0\). Each expression differentiates to \(x^{-3}\) on its interval.

The Exponent Negative One

The exceptional exponent deserves a direct check. If \(r=-1\), then the power-rule expression would require division by \(r+1=0\). Thus writing \(x^{r+1}/(r+1)\) is not a valid calculation. This is not a minor arithmetic inconvenience; it marks a different antiderivative rule.

Key takeaway: The power rule for antiderivatives applies only when the exponent is not \(-1\). For \(x^{-1}=1/x\), do not substitute \(r=-1\) into the power-rule formula; use the separate logarithmic antiderivative rule.

A Common Mistake: Forgetting the Reciprocal

A frequent error is to rewrite a reciprocal radical as a positive power. For example, \(1/\sqrt{x}\) is not \(x^{1/2}\); it is \(x^{-1/2}\). Applying the rule to the incorrect exponent changes the derivative and therefore cannot give an antiderivative of the original integrand.

$$ \text{Incorrect: }\quad \int \frac{1}{\sqrt{x}}\,dx \ne \frac{x^{3/2}}{3/2}+C. $$

Indeed, differentiating the incorrect proposed answer gives \(\sqrt{x}\), not \(1/\sqrt{x}\). The correction is to place the power in the denominator before converting to an exponent:

$$ \frac{1}{\sqrt{x}}=x^{-1/2} \quad\Longrightarrow\quad \int \frac{1}{\sqrt{x}}\,dx=2\sqrt{x}+C,\qquad x>0. $$

Always check the rewrite against the original expression, apply the exponent rule only after confirming its condition, and differentiate the result to verify it.

Check Your Understanding

For each question, identify the domain where needed and use differentiation to check any proposed antiderivative.

  1. Rewrite \(1/\sqrt[3]{x^2}\) using a negative fractional exponent. What restriction does the denominator impose?
  2. Find an antiderivative of \(x^{-2/3}\) on \(x>0\).
  3. Find an antiderivative of \(\sqrt[3]{x}\) on the real line.
  4. Why can the power rule not be applied to \(x^{-1}\)?
  5. On which intervals is \(x^{-4}\) defined, and may its antiderivative use different constants on those intervals?

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