AP Calculus BC

Integral of arctan x: Answer, Parts Setup, Mistakes

Q: What is the integral of arctan x?

It is x\arctan x - \frac{1}{2}\ln\left(1 + x^{2}\right) + C. Integration by parts with u = \arctan x and dv = dx gives x\arctan x - \int \frac{x}{1 + x^{2}}\,dx, and the substitution w = 1 + x^{2} finishes it.

Q: Why is there no absolute value around 1 + x^2?

Because 1 + x^{2} is at least 1 for every real x, so it is never zero or negative. Absolute value bars exist to keep the logarithm defined, and here there is nothing to protect against.

Q: What is the definite integral of arctan x from 0 to 1?

It equals \frac{\pi}{4} - \frac{\ln 2}{2}, about 0.439. The lower endpoint contributes nothing because \arctan 0 = 0 and \ln 1 = 0.

Q: How do I integrate arctan of a multiple, like arctan(3x)?

Substitute w = 3x first, so dx = \frac{dw}{3} and the integral becomes \frac{1}{3}\int \arctan w\,dw = \frac{1}{3}\left(w\arctan w - \frac{1}{2}\ln\left(1 + w^{2}\right)\right) + C, then replace w with 3x.

The integral of arctan x with respect to x is x arctan x - (1/2) ln(1 + x^2) + C. Integration by parts with u = arctan x and dv = dx gives x arctan x minus the integral of x over 1 + x^2, and that leftover integral is a substitution with u = 1 + x^2. No absolute value is needed because 1 + x^2 is always positive.

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

How to integrate arctan x by parts

As with $\ln x$ and $\arcsin x$ , there is only one visible factor, so the $dx$ becomes the second one. Write the integrand as $\arctan x \cdot 1$ and antidifferentiate the $1$ .

\int u\,dv = uv - \int v\,du

Take $u = \arctan x$ , whose derivative $\frac{1}{1 + x^{2}}$ is purely algebraic, and $dv = dx$ , so $v = x$ .

u = \arctan x, \quad du = \frac{1}{1 + x^{2}}\,dx, \quad dv = dx, \quad v = x

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

The leftover integral is a substitution: let $w = 1 + x^{2}$ , so $dw = 2x\,dx$ and $x\,dx = \frac{1}{2}\,dw$ .

\int \frac{x}{1 + x^{2}}\,dx = \frac{1}{2}\int \frac{1}{w}\,dw = \frac{1}{2}\ln\left(1 + x^{2}\right)

Combining the pieces, the minus sign from the parts formula stays, because this substitution introduced no second negative.

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

Check it by differentiating

The product rule on $x\arctan x$ gives $\arctan x + \frac{x}{1 + x^{2}}$ , and differentiating $-\frac{1}{2}\ln\left(1 + x^{2}\right)$ gives $-\frac{1}{2}\cdot\frac{2x}{1 + x^{2}} = -\frac{x}{1 + x^{2}}$ . The fractions cancel, leaving $\arctan x$ .

Why there is no absolute value in the logarithm

Most antiderivatives that produce a logarithm carry bars, as in $\int \frac{1}{x}\,dx = \ln\left|x\right| + C$ , because the argument can be negative. Here it cannot: $1 + x^{2} \ge 1$ for every real $x$ , so the argument is always positive and the bars would be decoration.

Writing $\ln\left|1 + x^{2}\right|$ is not wrong, just redundant. It is worth understanding the reason rather than memorizing which results take bars, because the rule is simply whether the argument can reach zero or below.

It is also worth keeping the two arctangent facts apart, since they point in opposite directions and are easy to swap under time pressure.

$\frac{d}{dx}\arctan x = \frac{1}{1 + x^{2}}$ , so $\int \frac{1}{1 + x^{2}}\,dx = \arctan x + C$
$\int \arctan x\,dx = x\arctan x - \frac{1}{2}\ln\left(1 + x^{2}\right) + C$ , which needs parts

The first is a basic antiderivative on both AB and BC. The second is a BC parts problem. Confusing them is the single most common error on this topic.

Where the integral of arctan x shows up on the AP exam

Integration by parts is Topic 6.11, marked BC only in the CED, so this integral is BC material. Unit 6 carries a weighting of 15 to 20 percent on both exams. The derivative of $\arctan x$ comes earlier, in Topic 3.4 (Differentiating Inverse Trigonometric Functions), which is on both exams.

Definite versions from $0$ are the usual form, because $\arctan 0 = 0$ and $\ln 1 = 0$ make the lower endpoint vanish entirely.

\int_{0}^{1} \arctan x\,dx = \left[x\arctan x - \frac{1}{2}\ln\left(1 + x^{2}\right)\right]_{0}^{1} = \frac{\pi}{4} - \frac{\ln 2}{2}

That is about $0.439$ . At $x = 1$ the first term is $\arctan 1 = \frac{\pi}{4}$ and the second is $\frac{1}{2}\ln 2$ ; at $x = 0$ both terms are zero.

The related Maclaurin series is separate BC material. Term by term integration of $\arctan x = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \cdots$ is a standard series question, and it converges on $-1 \le x \le 1$ .

Common mistakes with the integral of arctan x

Answering $\frac{1}{1 + x^{2}}$ . That is the derivative of $\arctan x$ , not its antiderivative, and the two are asked in nearly identical wording.
Writing $\frac{\left(\arctan x\right)^{2}}{2}$ . The power rule does not apply to a function of $x$ unless its derivative is also present in the integrand.
Losing the $\frac{1}{2}$ on the logarithm. The substitution $w = 1 + x^{2}$ gives $dw = 2x\,dx$ , and that $2$ has to be paid for.
Flipping the sign to $+\frac{1}{2}\ln\left(1 + x^{2}\right)$ . Unlike the $\arcsin x$ case, no second minus appears here, so the parts minus survives to the final answer.
Using degrees. $\arctan$ in calculus is always in radians, so $\arctan 1$ is $\frac{\pi}{4}$ , not $45$ .

The sign differs from the arcsin case for a reason

For $\arcsin x$ the leftover substitution produced a negative, which cancelled the parts minus and gave a plus. For $\arctan x$ the substitution produces a positive, so the parts minus stays. Tracking where each sign comes from is more reliable than memorizing both results.

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 arctan x?

It is $x\arctan x - \frac{1}{2}\ln\left(1 + x^{2}\right) + C$ . Integration by parts with $u = \arctan x$ and $dv = dx$ gives $x\arctan x - \int \frac{x}{1 + x^{2}}\,dx$ , and the substitution $w = 1 + x^{2}$ finishes it.

Is the integral of arctan x on the AP Calculus AB exam?

No. It requires integration by parts, Topic 6.11, which the CED marks BC only. Both exams do cover $\int \frac{1}{1 + x^{2}}\,dx = \arctan x + C$ , which is the opposite direction.

Why is there no absolute value around 1 + x^2?

Because $1 + x^{2}$ is at least $1$ for every real $x$ , so it is never zero or negative. Absolute value bars exist to keep the logarithm defined, and here there is nothing to protect against.

What is the definite integral of arctan x from 0 to 1?

It equals $\frac{\pi}{4} - \frac{\ln 2}{2}$ , about $0.439$ . The lower endpoint contributes nothing because $\arctan 0 = 0$ and $\ln 1 = 0$ .

How do I integrate arctan of a multiple, like arctan(3x)?

Substitute $w = 3x$ first, so $dx = \frac{dw}{3}$ and the integral becomes $\frac{1}{3}\int \arctan w\,dw = \frac{1}{3}\left(w\arctan w - \frac{1}{2}\ln\left(1 + w^{2}\right)\right) + C$ , then replace $w$ with $3x$ .