Derivative of an accumulation function at a point

By FTC Part 1, since $F(x) = \int_0^x \arctan(t)\, dt$ with continuous integrand $f(t) = \arctan(t)$, we know $F'(x) = f(x) = \arctan(x)$ for all $x$. Note the upper limit here is just $x$ itself, so no chain rule adjustment is needed.

Substituting $x = 1$ gives $F'(1) = \arctan(1)$. Since $\arctan(1) = \pi/4$, we conclude $F'(1) = \pi/4$.