AP Calculus AB and BC

Integral of 1/(9+x^2): Answer, Proof, and Steps

Q: How do you know to use arctangent here?

Because the denominator is a sum of squares, a^2 + x^2. That pattern always integrates to \frac{1}{a}\arctan\frac{x}{a} + C. Here a = 3.

The integral of 1/(9+x^2) is (1/3)arctan(x/3) + C. This matches the standard arctangent form with a = 3, since 9 = 3^2. Differentiating (1/3)arctan(x/3) simplifies back to 1/(9+x^2), which confirms the answer.

\int \frac{1}{9+x^2}\,dx = \frac{1}{3}\arctan\frac{x}{3} + C

Matching the arctangent form

A constant over a sum of squares integrates to an arctangent. The standard form has $a^2 + x^2$ in the denominator.

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

Here $9 = 3^2$ , so $a = 3$ . Substitute directly into the formula.

\int \frac{1}{9 + x^2}\,dx = \int \frac{1}{3^2 + x^2}\,dx = \frac{1}{3}\arctan\frac{x}{3} + C

Read off a from the constant

The constant term is $a^2$ , so take its square root to find $a$ . Here $a = \sqrt{9} = 3$ .

Verifying by differentiating

Differentiate the result. The chain rule supplies an inner factor of $\frac{1}{3}$ from the argument $\frac{x}{3}$ .

\frac{d}{dx}\left(\frac{1}{3}\arctan\frac{x}{3}\right) = \frac{1}{3}\cdot\frac{1}{1 + \left(\frac{x}{3}\right)^2}\cdot\frac{1}{3} = \frac{1}{9}\cdot\frac{1}{1 + \frac{x^2}{9}}

Multiplying the top and bottom of the last fraction by $9$ clears the inner fraction and returns the integrand.

\frac{1}{9}\cdot\frac{1}{1 + \frac{x^2}{9}} = \frac{1}{9 + x^2}

Common mistakes

Forgetting the $\frac{1}{a}$ factor. The answer is $\frac{1}{3}\arctan\frac{x}{3}$ , not $\arctan\frac{x}{3}$ .
Using $a = 9$ instead of $a = 3$ . The constant is $a^2 = 9$ , so $a = 3$ .
Confusing this with the arcsine form. A plain sum $9 + x^2$ gives arctangent; a difference under a root like $\sqrt{9 - x^2}$ gives arcsine.

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/(9+x^2) from 0 to 3?

$\int_0^3 \frac{1}{9+x^2}\,dx = \left[\frac{1}{3}\arctan\frac{x}{3}\right]_0^3 = \frac{1}{3}\arctan 1 = \frac{1}{3}\cdot\frac{\pi}{4} = \frac{\pi}{12}$ .

How do you know to use arctangent here?

Because the denominator is a sum of squares, $a^2 + x^2$ . That pattern always integrates to $\frac{1}{a}\arctan\frac{x}{a} + C$ . Here $a = 3$ .

What is the general integral of 1/(a^2+x^2)?

$\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C$ for any $a > 0$ . The $\frac{1}{a}$ out front comes from the chain rule on $\arctan\frac{x}{a}$ .