AP Calculus BC

Integral of 1/(x^2-1): Partial Fractions

The integral of 1 over x squared minus 1 is one half times the natural log of the absolute value of x minus 1 over x plus 1, plus C. The denominator factors, so partial fractions splits it into two simple logarithms.

∫1x2−1 dx=12ln⁡∣x−1x+1∣+C\int \frac{1}{x^{2}-1}\,dx = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C

Split the denominator

1x2−1=1(x−1)(x+1)=1/2x−1−1/2x+1\frac{1}{x^{2}-1} = \frac{1}{(x-1)(x+1)} = \frac{1/2}{x-1} - \frac{1/2}{x+1}
∫=12ln⁡∣x−1∣−12ln⁡∣x+1∣+C=12ln⁡∣x−1x+1∣+C\int = \frac{1}{2}\ln\left|x-1\right| - \frac{1}{2}\ln\left|x+1\right| + C = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C

Factoring first is the whole decision

A quadratic denominator that FACTORS wants partial fractions. One that does not factor wants completing the square and an arctangent. Checking the discriminant picks the technique before any work begins.

Three separate intervals

The integrand has asymptotes at x=1x = 1 and x=−1x = -1, so the antiderivative is valid on each of the three intervals separately and no definite integral may cross either point.

Integrating using linear partial fractions is Topic 6.12, which the CED marks BC only.

Common mistakes

  • Answering ln⁡∣x2−1∣\ln\left|x^{2}-1\right|. The numerator would have to be 2x2x for that.
  • Getting the constants wrong. Both are 12\frac{1}{2} in size, and the second is subtracted.
  • Integrating across x=1x = 1 or 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^2-1)?

It is 12ln⁡∣x−1x+1∣+C\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C.

How do I know to use partial fractions?

The denominator factors over the reals. If it did not, you would complete the square and get an arctangent.