Integration by substitution

Define $\Phi(x) = F(g(x))$. By the chain rule, since $F'=f$, $\Phi'(x) = F'(g(x))\cdot g'(x) = f(g(x))g'(x)$.

This says exactly that $\Phi$ is differentiable with derivative $f(g(x))g'(x)$ at every $x$ in the domain — i.e. $\Phi$ is, by definition, an antiderivative of $f(g(x))g'(x)$.

By the uniqueness theorem above, every antiderivative of $f(g(x))g'(x)$ differs from $\Phi$ by a constant, so $\int f(g(x))g'(x)\,dx = \Phi(x) + C = F(g(x)) + C$, as claimed. In practice one writes $u=g(x)$, $du=g'(x)\,dx$, reducing the left side to $\int f(u)\,du = F(u)+C$ before substituting $u=g(x)$ back at the end.