> [!NOTE] Theorem > Let $\frac{d}{dt} x(t) = f(x(t))g(t), \quad x(t_{0})=x_{0}\tag{1}$with $t\in(\alpha,\beta)\subset \mathbb{R},$ $f:\mathbb{R}\to \mathbb{R}$ and $g:(\alpha,\beta)\to \mathbb{R}$ be an [[Initial Value Problem for Scalar Ordinary Differential Equation|initial value problem]] for a [[Separable Differential Equation|separable differential equation]] that has a unique [[Solution to Scalar Ordinary Differential Equation|solution]] and $f(x(t))\neq 0$ for all $t\in(\alpha,\beta).$ > >Then $\int_{x_{0}}^{x(t)} \frac{1}{f(\tilde{x})} \, d\tilde{x} = \int_{t_{0}}^{t} g(\tilde{t}) \, d\tilde{t}$for all $t\in(\alpha,\beta).$ **Proof**: dividing both sides of $(1)$ by $f(x(t))$ gives $\frac{1}{f(x(t))} \frac{d}{dt}x(t)=g(t)$Let $F(\tilde{x})$ denote an anti-derivative of $1/f(\tilde{x}),$ so that $F'(\tilde{x})= 1/f(\tilde{x}).$ By chain rule $\frac{d}{dt}F(x(t))=F'(x(t)) \frac{d}{dt}x(t) = \frac{1}{f(x(t))} \frac{d}{dt}x(t)$Substituting gives $\frac{d}{dt} F(x(t)) = g(t)$Integrating with respect to $t$ and using the initial condition gives $F(x(t))-F(x_{0})=\int_{t_{0}}^{t} g(\tilde{t}) \, d\tilde{t} $