AP Calculus AB and BC

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

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

It is \arctan x + C. The result follows from reversing the derivative rule \frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}, so it is a basic antiderivative you recall rather than derive during the exam.

Q: Why is the answer arctan and not a logarithm?

A logarithm appears when the numerator is the derivative of the denominator. Here the denominator 1+x^2 has derivative 2x, and the numerator is 1, so u-substitution fails. Since 1+x^2 never factors over the reals, the antiderivative is the inverse tangent instead.

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

\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + C for a > 0. For example, \int \frac{1}{16+x^2}\,dx = \frac{1}{4}\arctan\!\left(\frac{x}{4}\right) + C. The \frac{1}{a} out front is the part students most often lose.

Q: Is this integral on the AP Calculus AB exam or only BC?

Both. The derivative of \arctan x is Topic 3.4 and the matching antiderivative is Topic 6.8, and neither is marked BC only. BC adds the improper version, where \int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx = \pi.

The integral of 1/(1+x^2) with respect to x is arctan(x) + C. It is the reverse of the derivative rule d/dx[arctan x] = 1/(1+x^2), so it is a memorized antiderivative, not a substitution problem. The general form is the integral of 1/(a^2+x^2), which equals (1/a)arctan(x/a) + C for a > 0.

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

Why the antiderivative is arctan x

This integral is not solved, it is recognized. Every basic antiderivative is a derivative rule read backwards, and $\frac{1}{1+x^2}$ is exactly what you get when you differentiate the inverse tangent.

\frac{d}{dx}\left[\arctan x\right] = \frac{1}{1+x^2}

Reversing that statement gives the indefinite integral, with the constant of integration attached because any two antiderivatives of the same function differ by a constant.

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

You can confirm the derivative rule itself with implicit differentiation. Write $y = \arctan x$ , which is the same as $\tan y = x$ for $-\frac{\pi}{2} < y < \frac{\pi}{2}$ , then differentiate both sides with respect to $x$ .

\sec^2 y \cdot \frac{dy}{dx} = 1 \quad \Longrightarrow \quad \frac{dy}{dx} = \frac{1}{\sec^2 y}

The Pythagorean identity $\sec^2 y = 1 + \tan^2 y$ turns the answer back into a function of $x$ , since $\tan y = x$ .

\frac{dy}{dx} = \frac{1}{1+\tan^2 y} = \frac{1}{1+x^2}

The domain is every real number

The denominator $1+x^2$ is at least 1, so it never vanishes. Both $\frac{1}{1+x^2}$ and $\arctan x$ are defined for all real $x$ , which means this integrand has no vertical asymptotes and no improper behavior on any finite interval. That is unusual for a rational function and it is part of why the arctangent shows up so cleanly.

The general a^2 + x^2 form

Most exam versions of this integral do not arrive with a bare 1 in the denominator. The pattern to learn is the one with a constant, where the answer picks up a factor of $\frac{1}{a}$ in front and divides the argument by $a$ .

\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + C, \quad a > 0

Example 1. Integrate $\frac{1}{9+x^2}$ . Here $a^2 = 9$ , so $a = 3$ .

\int \frac{1}{9+x^2}\,dx = \frac{1}{3}\arctan\!\left(\frac{x}{3}\right) + C

Example 2. Integrate $\frac{1}{1+4x^2}$ . Rewrite $4x^2$ as $(2x)^2$ and substitute $u = 2x$ , so $du = 2\,dx$ and $dx = \frac{du}{2}$ .

\int \frac{1}{1+4x^2}\,dx = \frac{1}{2}\int \frac{1}{1+u^2}\,du = \frac{1}{2}\arctan(2x) + C

Example 3. Integrate $\frac{2x}{1+x^4}$ . The numerator is the derivative of $x^2$ , so $u = x^2$ collapses this to the base case.

\int \frac{2x}{1+x^4}\,dx = \int \frac{1}{1+u^2}\,du = \arctan(x^2) + C

Example 4. A definite integral. Because $\arctan 1 = \frac{\pi}{4}$ and $\arctan 0 = 0$ , the Fundamental Theorem of Calculus gives a clean exact value.

\int_0^1 \frac{1}{1+x^2}\,dx = \left[\arctan x\right]_0^1 = \frac{\pi}{4} - 0 = \frac{\pi}{4}

Where this integral shows up on the AP exam

Two CED topics own this result. Topic 3.4, Differentiating Inverse Trigonometric Functions, supplies the derivative $\frac{1}{1+x^2}$ , and Topic 6.8, Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation, is where you are expected to read it backwards. Unit 3 carries a weighting of 5 to 10 percent on both AB and BC; Unit 6 carries 15 to 20 percent on both.

When the integrand has a constant inside, or a chain of the form $\frac{u'}{1+u^2}$ , the work belongs to Topic 6.9, Integrating Using Substitution. On BC, the same antiderivative reappears in improper integrals, where the exact value of the whole real line is finite.

\int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx = \lim_{b \to \infty}\arctan b - \lim_{a \to -\infty}\arctan a = \frac{\pi}{2} + \frac{\pi}{2} = \pi

Recognition beats technique here

On a no-calculator free-response part, the fastest correct path is to name the form. If the denominator is a constant plus a square and the numerator is a constant, reach for the arctangent before you try anything else. Scanning for a technique you do not need is where the clock goes.

Common mistakes with this integral

Reaching for partial fractions. Topic 6.12 covers linear partial fractions, and $1+x^2$ does not factor over the real numbers, so there is nothing to decompose. The arctangent is the answer precisely because the quadratic is irreducible.
Confusing it with $\frac{x}{1+x^2}$ . That numerator makes $u = 1+x^2$ work and the answer is a logarithm: $\int \frac{x}{1+x^2}\,dx = \frac{1}{2}\ln(1+x^2) + C$ . One stray factor of $x$ changes the whole family of functions.
Confusing it with $\frac{1}{\sqrt{1-x^2}}$ . The square root and the minus sign give arcsine instead: $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C$ . Check whether the quadratic is under a radical and whether the sign is plus or minus before you commit.
Dropping the $\frac{1}{a}$ in the general form. $\int \frac{1}{9+x^2}\,dx$ is $\frac{1}{3}\arctan\!\left(\frac{x}{3}\right) + C$ , not $\arctan\!\left(\frac{x}{3}\right) + C$ . Differentiate your answer to catch this in one line.
Writing $\ln\lvert 1+x^2\rvert$ because the integrand is one over something. The logarithm rule needs the derivative of the denominator on top, and the derivative of $1+x^2$ is $2x$ , not 1.
Leaving off $+ C$ on the indefinite integral. On a differential-equation problem the missing constant is the difference between a general solution and no solution at all.

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^2)?

It is $\arctan x + C$ . The result follows from reversing the derivative rule $\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$ , so it is a basic antiderivative you recall rather than derive during the exam.

Why is the answer arctan and not a logarithm?

A logarithm appears when the numerator is the derivative of the denominator. Here the denominator $1+x^2$ has derivative $2x$ , and the numerator is 1, so $u$ -substitution fails. Since $1+x^2$ never factors over the reals, the antiderivative is the inverse tangent instead.

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

$\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + C$ for $a > 0$ . For example, $\int \frac{1}{16+x^2}\,dx = \frac{1}{4}\arctan\!\left(\frac{x}{4}\right) + C$ . The $\frac{1}{a}$ out front is the part students most often lose.

Is this integral on the AP Calculus AB exam or only BC?

Both. The derivative of $\arctan x$ is Topic 3.4 and the matching antiderivative is Topic 6.8, and neither is marked BC only. BC adds the improper version, where $\int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx = \pi$ .

How do I check an antiderivative like this one?

Differentiate your answer and compare it to the original integrand. If $F(x) = \arctan x$ then $F'(x) = \frac{1}{1+x^2}$ , which matches, so $F$ is correct. This one-line check catches a missing $\frac{1}{a}$ or a bad chain-rule factor immediately.