Evaluating a definite integral using arctan(x) as antiderivative

Since $F(x) = \arctan(x)$ is an antiderivative of $f(x) = \dfrac{1}{1+x^2}$, FTC Part 2 tells us that $\int_0^1 \frac{1}{1+x^2}\,dx = \arctan(x)\Big|_0^1 = \arctan(1) - \arctan(0)$.

We know $\arctan(1) = \pi/4$ and $\arctan(0) = 0$, so the integral equals $\pi/4 - 0 = \pi/4$.

The answer is arctangent of zero. In the earlier part, the integral was evaluated as arctangent of one minus arctangent of zero, following FTC Part 2's rule of subtracting the antiderivative's value at the lower limit from its value at the upper limit. Since arctangent of zero equals zero, the final result simplified to pi over four.