AP Calculus AB and BC

Integral of arctan x/(1+x^2)

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

It is \frac{\left(\arctan x\right)^{2}}{2} + C.

Q: Which pattern is this?

\int u\,du = \frac{u^{2}}{2}, where a function is multiplied by its own derivative.

The integral of arctan x over 1 plus x squared is arctan x squared over 2, plus C. The 1 over 1 plus x squared is exactly the derivative of arctan x, so substituting u equals arctan x turns the whole thing into the integral of u.

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

The function and its own derivative

u = \arctan x, \quad du = \frac{dx}{1+x^{2}} \implies \int u\,du = \frac{u^{2}}{2} + C

This is the $\int u\,du$ pattern: whenever a function appears multiplied by its own derivative, the antiderivative is half its square.

The same shape recurs

It is the reason the integral of ln x over x is (ln x)^2/2, and of sin x cos x is sin^2 x over 2. One pattern, three standard results.

Common mistakes

Answering $\frac{\arctan x}{2}$ , forgetting to square.
Reaching for integration by parts. The substitution is immediate.
Reading $\left(\arctan x\right)^{2}$ as $\arctan\left(x^{2}\right)$ .

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)/(1+x^2)?

It is $\frac{\left(\arctan x\right)^{2}}{2} + C$ .

Which pattern is this?

$\int u\,du = \frac{u^{2}}{2}$ , where a function is multiplied by its own derivative.