How to evaluate an arctangent-type definite integral

Key concept: This integral is a familiar one: the derivative of $\arctan x$ is sitting right underneath the integrand. Once you spot that, the bounds do the rest of the work almost immediately.

Step by step

We use the standard antiderivative

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

So

=\arctan(\sqrt{3})-\arctan(1).$$ Now evaluate the special angles: $$\arctan(\sqrt{3})=\frac{\pi}{3}, \qquad \arctan(1)=\frac{\pi}{4}.$$ Therefore $$\frac{\pi}{3}-\frac{\pi}{4}=\frac{4\pi-3\pi}{12}=\frac{\pi}{12}.$$ ## Answer $$\boxed{\frac{\pi}{12}}$$ ### Pitfall alert The main mistake here is forgetting that the antiderivative is $\arctan x$, not $\tan^{-1}x$ in the sense of reciprocal. Another issue is mixing up the exact angles: $\arctan(\sqrt{3})$ is $\pi/3$, while $\arctan(1)$ is $\pi/4$. If you rush the subtraction, it is easy to get the denominator wrong. ### Try different conditions If the limits were reversed, the value would be $-\pi/12$. If the upper limit were $1$ and the lower limit were $0$, the integral would be $\arctan(1)-\arctan(0)=\pi/4$. The same antiderivative works for any interval where the integrand is defined. ### Further reading inverse tangent, definite integral, antiderivative