AP Calculus AB and BC

Integral of 1/(x (ln x)^2): Substitution

The integral of 1 over x times ln x squared is negative 1 over ln x, plus C. Substituting u equal to ln x turns it into the integral of u to the negative 2, and unlike the version with a single logarithm this one converges at infinity.

∫1x(ln⁡x)2 dx=−1ln⁡x+C\int \frac{1}{x\left(\ln x\right)^{2}}\,dx = -\frac{1}{\ln x} + C

The substitution

u=ln⁡x,du=dxx  ⟹  ∫u−2du=−1ln⁡x+Cu = \ln x, \quad du = \frac{dx}{x} \implies \int u^{-2}du = -\frac{1}{\ln x} + C

The exponent decides convergence

This is the boundary case that separates two famous results. With a single logarithm the integral diverges; with a square it converges.

∫2∞dxxln⁡x=∞,∫2∞dxx(ln⁡x)2=1ln⁡2\int_{2}^{\infty}\frac{dx}{x\ln x} = \infty, \qquad \int_{2}^{\infty}\frac{dx}{x\left(\ln x\right)^{2}} = \frac{1}{\ln 2}

By the integral test the matching series behave the same way, which is why ∑1nln⁡n\sum \frac{1}{n\ln n} diverges while ∑1n(ln⁡n)2\sum \frac{1}{n(\ln n)^{2}} converges.

Common mistakes

  • Answering a logarithm. That is the exponent −1-1 case; here the exponent is −2-2.
  • Starting at x=1x = 1, where ln⁡1=0\ln 1 = 0 makes the integrand undefined.

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Frequently asked questions

What is the integral of 1/(x (ln x)^2)?

It is −1ln⁡x+C-\frac{1}{\ln x} + C.

Does it converge at infinity?

Yes, to 1ln⁡2\frac{1}{\ln 2} from x=2x = 2. The single-logarithm version diverges.