AP Calculus AB and BC

Integral of 1/x^3: Answer, Power Rule, Mistakes

Q: What is the integral of 1/x^3?

It is -\frac{1}{2x^2} + C. Rewrite \frac{1}{x^3} as x^{-3}; the power rule raises the exponent to -2 and divides by -2, giving -\frac{1}{2}x^{-2}.

Q: Why isn't the integral of 1/x^3 a logarithm?

The logarithm appears only for the exponent -1, where the power rule would divide by zero. The exponent here is -3, so the ordinary power rule applies and the answer is a power, -\frac{1}{2}x^{-2}.

Q: Can I integrate 1/x^3 from -1 to 1?

No. The integrand has a vertical asymptote at x = 0 inside that interval, so the integral is improper and does not converge. Substituting the endpoints mechanically gives a meaningless value.

The integral of 1/x^3 with respect to x is -1/(2x^2) + C. Rewrite the integrand as x^-3 and apply the power rule, which raises the exponent to -2 and divides by -2. As with any negative power of x, you cannot integrate across x = 0, where the function is undefined.

\int \frac{1}{x^3}\,dx = -\frac{1}{2x^2} + C

How to integrate 1/x^3 with the power rule

Written as a fraction it looks like a quotient, and there is no quotient rule for integrals. Move the power upstairs first.

\frac{1}{x^3} = x^{-3}

Now the power rule for antiderivatives applies. Raise the exponent by one and divide by the new exponent; it holds for every exponent except $-1$ .

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1

With $n = -3$ , the new exponent is $-2$ and you divide by $-2$ .

\int x^{-3}\,dx = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C

Check it by differentiating

Differentiate $-\frac{1}{2}x^{-2}$ : multiply by the exponent $-2$ and lower it, giving $\left(-\frac{1}{2}\right)(-2)x^{-3} = x^{-3} = \frac{1}{x^3}$ . The factor of $\frac{1}{2}$ is exactly what the $-2$ cancels.

Where the integral of 1/x^3 shows up on the AP exam

This is a Topic 6.8 basic antiderivative, part of the 15 to 20 percent that Unit 6 carries. The only subtlety is the domain: $\frac{1}{x^3}$ has a vertical asymptote at $x = 0$ , so a definite integral is valid only on an interval that stays on one side of the origin.

\int_{1}^{2} \frac{1}{x^3}\,dx = \left[-\frac{1}{2x^2}\right]_{1}^{2} = -\frac{1}{8} + \frac{1}{2} = \frac{3}{8}

The improper integral out to infinity converges, since $\frac{1}{x^3}$ is a $p$ -integrand with $p = 3 > 1$ .

\int_{1}^{\infty} \frac{1}{x^3}\,dx = \lim_{b \to \infty}\left[-\frac{1}{2x^2}\right]_{1}^{b} = \frac{1}{2}

Common mistakes with the integral of 1/x^3

Answering a logarithm. The $\ln$ form belongs only to $\frac{1}{x}$ , where the exponent is $-1$ . With exponent $-3$ the power rule works normally.
Forgetting the $\frac{1}{2}$ . Dividing by the new exponent $-2$ produces it; writing $-\frac{1}{x^2}$ differentiates to $\frac{2}{x^3}$ , twice too big.
Sign errors. The answer is negative, $-\frac{1}{2x^2}$ ; a positive $\frac{1}{2x^2}$ differentiates to $-\frac{1}{x^3}$ , the wrong sign.
Integrating across $x = 0$ . On an interval like $[-1, 1]$ the integrand blows up at the origin, so the integral is improper and mechanical endpoint substitution is meaningless.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of 1/x^3?

It is $-\frac{1}{2x^2} + C$ . Rewrite $\frac{1}{x^3}$ as $x^{-3}$ ; the power rule raises the exponent to $-2$ and divides by $-2$ , giving $-\frac{1}{2}x^{-2}$ .

Why isn't the integral of 1/x^3 a logarithm?

The logarithm appears only for the exponent $-1$ , where the power rule would divide by zero. The exponent here is $-3$ , so the ordinary power rule applies and the answer is a power, $-\frac{1}{2}x^{-2}$ .

Can I integrate 1/x^3 from -1 to 1?

No. The integrand has a vertical asymptote at $x = 0$ inside that interval, so the integral is improper and does not converge. Substituting the endpoints mechanically gives a meaningless value.