AP Calculus BC

Integral of x arctan x: Parts Then Division

Q: What is the integral of x arctan x?

It is \frac{\left(x^{2}+1\right)\arctan x - x}{2} + C.

Q: Why choose v = (x^2+1)/2?

Because v is any antiderivative of dv, and that choice makes the leftover integrand cancel to 1 instead of needing long division.

The integral of x arctan x is x squared plus 1 times arctan x, minus x, all over 2, plus C. Integration by parts with u equal to arctan x leaves a rational integral that needs long division before it can be finished.

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

Parts, with a twist on v

Take $u = \arctan x$ and $dv = x\,dx$ . Choosing $v = \frac{x^{2}+1}{2}$ rather than $\frac{x^{2}}{2}$ is legal, since they differ by a constant, and it makes the leftover integral collapse.

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

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

Choosing v cleverly

With the ordinary v = x^2/2 the leftover integral is x^2 over 2(1+x^2), which needs long division. Adding the constant 1/2 to v makes the fraction cancel outright. Any antiderivative of dv is a valid v.

Where it appears

Integration by parts is Topic 6.11, BC only. This problem is a favourite because it combines parts with a rational integral and rewards recognising that $v$ is not unique.

Common mistakes

Assuming $v$ must be $\frac{x^{2}}{2}$ . Any antiderivative of $dv$ works, and the smart choice saves a division step.
Forgetting that $\frac{x^{2}}{x^{2}+1}$ is improper and needs rewriting as $1 - \frac{1}{x^{2}+1}$ .
Choosing $u = x$ , which requires integrating arctangent and makes the problem harder.

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 x arctan x?

It is $\frac{\left(x^{2}+1\right)\arctan x - x}{2} + C$ .

Why choose v = (x^2+1)/2?

Because $v$ is any antiderivative of $dv$ , and that choice makes the leftover integrand cancel to $1$ instead of needing long division.