AP Calculus AB and BC

Integral of x/(x^2-1): Substitution, Not Partials

Q: What is the integral of x/(x^2-1)?

It is \frac{1}{2}\ln\left|x^{2}-1\right| + C.

Q: Do I need partial fractions for x/(x^2-1)?

No. The numerator is half the derivative of the denominator, so u = x^{2}-1 finishes it. Partial fractions gives the same answer after more work.

Q: Why is there an absolute value inside the logarithm?

Because x^{2}-1 is negative on -1 < x < 1, and \int\frac{du}{u} = \ln\left|u\right| + C is the version that covers negative u.

The integral of x over x squared minus 1 is one half times the natural log of the absolute value of x squared minus 1, plus C. The numerator is half the derivative of the denominator, which makes this a substitution problem rather than a partial fractions problem.

\int \frac{x}{x^{2}-1}\,dx = \frac{1}{2}\ln\left|x^{2}-1\right| + C

The numerator gives it away

The denominator is $x^{2}-1$ and its derivative is $2x$ . The numerator is $x$ , exactly half of that, which is the signal to substitute.

u = x^{2}-1,\quad du = 2x\,dx,\quad x\,dx = \frac{du}{2}

\int\frac{x}{x^{2}-1}\,dx = \frac{1}{2}\int\frac{du}{u} = \frac{1}{2}\ln\left|u\right| + C = \frac{1}{2}\ln\left|x^{2}-1\right| + C

The absolute value bars are load bearing. On $-1 < x < 1$ the quantity $x^{2}-1$ is negative while the integrand itself is continuous and finite, so without the bars the antiderivative would be undefined exactly where the function is best behaved.

Why partial fractions is the slower road

Since $x^{2}-1 = \left(x-1\right)\left(x+1\right)$ , partial fractions is legal: $\frac{x}{x^{2}-1} = \frac{1/2}{x-1} + \frac{1/2}{x+1}$ , giving $\frac{1}{2}\ln\left|x-1\right| + \frac{1}{2}\ln\left|x+1\right| + C$ . Combine the logarithms and it is the same answer.

So the choice is about labour, not about right and wrong. Compare the numerator with the derivative of the denominator before anything else. If it matches up to a constant multiple, substitution ends in two lines. Only when it does not, as with $\int\frac{dx}{x^{2}-1}$ , is the partial fractions setup earning its keep.

One constant per interval

The integrand is undefined at x = 1 and x = -1, so its domain is three separate intervals. An antiderivative may carry a different constant on each of them, because they are not connected. Within any one interval, +C behaves the usual way.

The mistakes students make

The technique is short enough that almost every lost mark here is a missing factor or a missing pair of bars.

Writing $\ln\left|x^{2}-1\right|$ with no coefficient. Its derivative is $\frac{2x}{x^{2}-1}$ , twice the integrand.
Writing $\frac{1}{2}\ln\left(x^{2}-1\right)$ without bars, which is undefined on $-1 < x < 1$ even though the integrand is continuous there.
Confusing the problem with $\int\frac{dx}{x^{2}-1}$ and quoting $\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|$ . That is the answer with no $x$ in the numerator.

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 x/(x^2-1)?

It is $\frac{1}{2}\ln\left|x^{2}-1\right| + C$ .

Do I need partial fractions for x/(x^2-1)?

No. The numerator is half the derivative of the denominator, so $u = x^{2}-1$ finishes it. Partial fractions gives the same answer after more work.

Why is there an absolute value inside the logarithm?

Because $x^{2}-1$ is negative on $-1 < x < 1$ , and $\int\frac{du}{u} = \ln\left|u\right| + C$ is the version that covers negative $u$ .