AP Calculus BC

Integral of 1/(x^2-4): A Log, Not an Arctangent

Q: What is the integral of 1/(x^2-4)?

Because x^{2}-4 splits into (x-2)(x+2), the answer is a difference of logs, written compactly as \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C.

Q: Why is 1/(x^2+4) an arctangent but 1/(x^2-4) a log?

Because x^{2}-4 factors into real linear pieces and x^{2}+4 does not. Real factors give partial fractions and logs; an irreducible quadratic under a constant numerator gives an arctangent.

Q: Where does the 1/4 come from?

From the partial fractions coefficients. The roots are 2 and -2, which are 4 apart, and that gap becomes the \frac{1}{4} in front of both logs.

The integral of 1 over x squared minus 4 is one quarter times the natural log of the absolute value of x minus 2 over x plus 2, plus C. The minus sign lets the denominator factor into x minus 2 times x plus 2, which is what makes this a partial fractions problem and not an arctangent.

\int \frac{1}{x^{2}-4}\,dx = \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C

The denominator factors, so split it

$x^{2}-4$ is a difference of squares: $(x-2)(x+2)$ . Distinct linear factors mean partial fractions, so set $\frac{1}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}$ and clear denominators to get $1 = A(x+2) + B(x-2)$ .

Substituting $x = 2$ gives $4A = 1$ , so $A = \frac{1}{4}$ . Substituting $x = -2$ gives $-4B = 1$ , so $B = -\frac{1}{4}$ . The $4$ is the gap between the two roots, which is where the coefficient in the final answer comes from.

\int \frac{dx}{x^{2}-4} = \frac{1}{4}\ln\left|x-2\right| - \frac{1}{4}\ln\left|x+2\right| + C = \frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C

Compare it with 1/(x^2+4)

Change one sign and the problem changes species. $x^{2}+4$ has no real roots, so it does not factor over the reals and there is nothing for partial fractions to split. That integrand is an arctangent.

\int \frac{dx}{x^{2}+4} = \frac{1}{2}\arctan\frac{x}{2} + C

Check the denominator before choosing a method

For a denominator shaped like x squared plus or minus a constant, the minus case factors into real linear pieces and gives a logarithm, and the plus case is irreducible and gives an arctangent. For any other quadratic the signs tell you nothing: check the discriminant. Real roots mean partial fractions, no real roots mean completing the square and an arctangent.

The mistakes students make

Each of these is a one-symbol slip that lands on a clean but incorrect answer, which is why they survive to the end of a free response.

Reaching for the arctangent pattern and writing $\frac{1}{2}\arctan\frac{x}{2}$ . That is the answer to $\int\frac{dx}{x^{2}+4}$ , a different problem.
Losing the $\frac{1}{4}$ and answering $\ln\left|\frac{x-2}{x+2}\right|$ , whose derivative is $\frac{4}{x^{2}-4}$ , four times too big.
Guessing $\ln\left|x^{2}-4\right|$ from the shape of the fraction. Its derivative is $\frac{2x}{x^{2}-4}$ , and there is no $2x$ in the numerator to justify it.

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^2-4)?

Because $x^{2}-4$ splits into $(x-2)(x+2)$ , the answer is a difference of logs, written compactly as $\frac{1}{4}\ln\left|\frac{x-2}{x+2}\right| + C$ .

Why is 1/(x^2+4) an arctangent but 1/(x^2-4) a log?

Because $x^{2}-4$ factors into real linear pieces and $x^{2}+4$ does not. Real factors give partial fractions and logs; an irreducible quadratic under a constant numerator gives an arctangent.

Where does the 1/4 come from?

From the partial fractions coefficients. The roots are $2$ and $-2$ , which are $4$ apart, and that gap becomes the $\frac{1}{4}$ in front of both logs.