AP Calculus AB and BC

Integral of 1/x^2: Answer, Power Rule, Mistakes

The integral of 1/x^2 with respect to x is -1/x + C. Rewrite the integrand as x^-2 and apply the power rule, which raises the exponent to -1 and divides by -1. The most important warning is that you cannot integrate 1/x^2 across x = 0, because the function is undefined there and the integral diverges.

∫1x2 dx=−1x+C\int \frac{1}{x^{2}}\,dx = -\frac{1}{x} + C

How to integrate 1/x^2 with the power rule

The only real step is rewriting. As long as the integrand is written as a fraction it looks like a quotient rule problem, and there is no quotient rule for integrals. Move the power upstairs first.

1x2=x−2\frac{1}{x^{2}} = x^{-2}

Now the power rule for antiderivatives applies directly. It says to raise the exponent by one and divide by the new exponent, and it is valid for every exponent except −1-1.

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

With n=−2n = -2, the new exponent is −1-1 and you divide by −1-1, which is where the minus sign in the answer comes from.

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

Check it by differentiating

Differentiate −x−1-x^{-1} using the power rule: multiply by the exponent −1-1 and lower it to −2-2, giving (−1)(−1)x−2=x−2=1x2\left(-1\right)\left(-1\right)x^{-2} = x^{-2} = \frac{1}{x^{2}}. The two negatives make the result positive, as it must be, since 1x2\frac{1}{x^{2}} is never negative.

Why 1/x and 1/x^2 have completely different answers

These two integrands look like neighbours and behave nothing alike. The power rule excludes exactly one exponent, n=−1n = -1, because that is the case where n+1=0n + 1 = 0 and the rule would divide by zero.

  • ∫1x2 dx=∫x−2 dx=−1x+C\int \frac{1}{x^{2}}\,dx = \int x^{-2}\,dx = -\frac{1}{x} + C, an ordinary power rule case
  • ∫1x dx=ln⁡∣x∣+C\int \frac{1}{x}\,dx = \ln\left|x\right| + C, the one exception, which is not a power at all

The behaviour at infinity differs just as sharply, and this is the fact BC students are expected to know cold. One improper integral converges and the other does not.

∫1∞1x2 dx=lim⁡b→∞[−1x]1b=1\int_{1}^{\infty} \frac{1}{x^{2}}\,dx = \lim_{b \to \infty}\left[-\frac{1}{x}\right]_{1}^{b} = 1
∫1∞1x dx=lim⁡b→∞[ln⁡x]1b=∞\int_{1}^{\infty} \frac{1}{x}\,dx = \lim_{b \to \infty}\left[\ln x\right]_{1}^{b} = \infty

The general statement is the p-integral rule: ∫1∞1xp dx\int_{1}^{\infty} \frac{1}{x^{p}}\,dx converges when p>1p > 1 and diverges when p≤1p \le 1. Here p=2p = 2 converges, and p=1p = 1 is the boundary that fails.

Where the integral of 1/x^2 shows up on the AP exam

Basic antiderivatives are Topic 6.8, on both AB and BC, and Unit 6 carries a weighting of 15 to 20 percent on both exams. Evaluating improper integrals is Topic 6.13, which the CED marks BC only, so the infinite versions above are BC material.

An ordinary definite integral on an interval that avoids the origin is routine.

∫141x2 dx=[−1x]14=−14+1=34\int_{1}^{4} \frac{1}{x^{2}}\,dx = \left[-\frac{1}{x}\right]_{1}^{4} = -\frac{1}{4} + 1 = \frac{3}{4}

The integrand also appears constantly in disguise. Inverse square relationships in related rates and in physics contexts are the same antiderivative wearing different letters.

∫kx2 dx=−kx+C\int \frac{k}{x^{2}}\,dx = -\frac{k}{x} + C

On BC, the convergence of ∫1∞1x2 dx\int_{1}^{\infty} \frac{1}{x^{2}}\,dx is also the integral test justification that ∑1n2\sum \frac{1}{n^{2}} converges, which is the p-series with p=2p = 2.

Common mistakes with the integral of 1/x^2

  • Writing ln⁡∣x2∣\ln\left|x^{2}\right| or ln⁡(x2)\ln\left(x^{2}\right). The logarithm answer belongs to 1x\frac{1}{x} alone. Differentiating ln⁡(x2)\ln\left(x^{2}\right) gives 2x\frac{2}{x}, nowhere near the integrand.
  • Losing the minus sign and writing 1x+C\frac{1}{x} + C. Differentiating that gives −1x2-\frac{1}{x^{2}}, which is the wrong sign everywhere.
  • Integrating across x=0x = 0. The classic trap is ∫−111x2 dx\int_{-1}^{1} \frac{1}{x^{2}}\,dx, where mechanical substitution gives −1−1=−2-1 - 1 = -2. A negative answer is impossible for a positive integrand, and the truth is that the integral diverges.
  • Applying the quotient rule pattern. There is no quotient rule for antiderivatives; rewriting as x−2x^{-2} is what makes the problem elementary.
  • Mixing up the p-rules. For INTEGRALS from 11 to infinity, p>1p > 1 converges. It is the same threshold as the p-series, so ∑1n2\sum \frac{1}{n^{2}} converging and ∑1n\sum \frac{1}{n} diverging match the integrals exactly.

A positive integrand cannot have a negative integral

If you ever compute a definite integral of 1x2\frac{1}{x^{2}} and get a negative number, you have integrated across a vertical asymptote. Check the interval for x=0x = 0 before you check your arithmetic.

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^2?

It is −1x+C-\frac{1}{x} + C. Rewrite 1x2\frac{1}{x^{2}} as x−2x^{-2}, then the power rule raises the exponent to −1-1 and divides by −1-1, producing the minus sign.

Why is the integral of 1/x^2 not a logarithm?

The logarithm appears only for the exponent −1-1, where the power rule would divide by zero. With exponent −2-2 the power rule works normally, so the answer is an ordinary power, −x−1-x^{-1}.

What is the integral of 1/x^2 from -1 to 1?

It does not exist. The integrand has a vertical asymptote at x=0x = 0 inside the interval, so the integral is improper and diverges. Substituting endpoints mechanically gives −2-2, which is impossible because the integrand is always positive.

Does the integral of 1/x^2 from 1 to infinity converge?

Yes, and it equals exactly 11. Since −1x→0-\frac{1}{x} \to 0 as x→∞x \to \infty, the value is 0−(−1)=10 - \left(-1\right) = 1. This is the p-integral with p=2>1p = 2 > 1.

Is the integral of 1/x^2 on the AP Calculus AB exam?

Yes. The antiderivative itself is Topic 6.8 basic material on both exams. Only the improper versions, taken to infinity or across the asymptote, are BC only, under Topic 6.13.