AP Calculus AB and BC

Integral of 1/(x-1): Answer and Bars

The integral of 1 over x minus 1 is the natural log of the absolute value of x minus 1, plus C. The substitution u equals x minus 1 has du equal to dx, so no compensating factor appears. The bars matter because x minus 1 is negative for x below 1.

∫1x−1 dx=ln⁡∣x−1∣+C\int \frac{1}{x-1}\,dx = \ln\left|x-1\right| + C

A shift costs nothing

With u=x−1u = x-1 the differential is du=dxdu = dx exactly, so the integral is the basic logarithm form with no factor to compensate for.

∫dxx−1=∫duu=ln⁡∣u∣+C=ln⁡∣x−1∣+C\int \frac{dx}{x-1} = \int \frac{du}{u} = \ln\left|u\right| + C = \ln\left|x-1\right| + C

Why the bars are not optional

The integrand is defined on both sides of x = 1, but ln of a negative number is not. The absolute value is what lets one formula cover both intervals.

You cannot integrate across x = 1

There is a vertical asymptote at x=1x = 1, so a definite integral whose interval contains 11 is improper and diverges. Mechanically substituting endpoints across it produces a finite-looking number that means nothing.

Common mistakes

  • Dropping the bars and writing ln⁡(x−1)\ln(x-1), which is undefined for x<1x < 1.
  • Writing ln⁡∣x∣−1\ln\left|x\right| - 1. The shift is inside the logarithm.
  • Integrating across the asymptote at x=1x = 1.

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-1)?

It is ln⁡∣x−1∣+C\ln\left|x-1\right| + C.

Why is there no factor out front?

Because u=x−1u = x-1 gives du=dxdu = dx with no constant. A coefficient like 2x−12x-1 would produce a 12\frac{1}{2}.

Can I integrate from 0 to 2?

No. The asymptote at x=1x = 1 sits inside that interval, so the integral is improper and divergent.