AP Calculus AB and BC

Integral of 1/(1-x): Watch the Sign

The integral of 1 over 1 minus x is negative the natural log of the absolute value of 1 minus x, plus C. The substitution u equals 1 minus x gives du equal to negative dx, and that minus sign survives into the answer.

∫11−x dx=−ln⁡∣1−x∣+C\int \frac{1}{1-x}\,dx = -\ln\left|1-x\right| + C

The negative inner derivative

u=1−x,du=−dx  ⟹  ∫dx1−x=−∫duu=−ln⁡∣1−x∣+Cu = 1-x, \quad du = -dx \implies \int \frac{dx}{1-x} = -\int \frac{du}{u} = -\ln\left|1-x\right| + C

Compare ∫dxx−1=+ln⁡∣x−1∣+C\int \frac{dx}{x-1} = +\ln\left|x-1\right| + C. The two integrands are negatives of each other, so the antiderivatives are too.

The geometric series link

On ∣x∣<1|x| < 1 the integrand is the sum of the geometric series 1+x+x2+⋯1 + x + x^{2} + \cdots, and integrating term by term gives the Maclaurin series for −ln⁡(1−x)-\ln(1-x), a standard BC result.

−ln⁡(1−x)=x+x22+x33+⋯ ,∣x∣<1-\ln(1-x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \cdots, \qquad |x| < 1

Common mistakes

  • Dropping the minus sign, which gives the antiderivative of 1x−1\frac{1}{x-1} instead.
  • Dropping the bars, which restricts the answer to x<1x < 1 only.

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/(1-x)?

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

Why the minus sign?

The substitution u=1−xu = 1-x gives du=−dxdu = -dx, and that minus carries through.

How does it relate to 1/(x-1)?

The integrands are negatives of each other, so their antiderivatives are too.