# Primitive of Composite Function

## Theorem

Let $f$ and $g$ be a real functions which are integrable.

Let the composite function $g \circ f$ also be integrable.

Then:

$\displaystyle \int g \circ f \left({x}\right) \ \mathrm d x = \int \frac {g \left({u}\right)}{f' \left({x}\right)} \ \mathrm d u$

where $u = f \left({x}\right)$.

## Proof

 $\displaystyle F \left({x}\right)$ $=$ $\displaystyle \int g \circ f \left({x}\right) \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \int g \left({f \left({x}\right)}\right) \ \mathrm d x$ Definition of Composite Function $\displaystyle$ $=$ $\displaystyle \int g \left({u}\right) \ \mathrm d x$ where $u = f \left({x}\right)$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d F} {\mathrm d x}$ $=$ $\displaystyle g \left({u}\right)$ Definition of Primitive $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d F} {\mathrm d u} \frac {\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle g \left({u}\right)$ Chain Rule for Derivatives $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d F} {\mathrm d u} \frac {\mathrm d} {\mathrm d x} f \left({x}\right)$ $=$ $\displaystyle g \left({u}\right)$ Definition of $u$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d F} {\mathrm d u} f' \left({x}\right)$ $=$ $\displaystyle g \left({u}\right)$ Definition of Derivative $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d F} {\mathrm d u}$ $=$ $\displaystyle \frac {g \left({u}\right)} {f' \left({x}\right)}$ $\displaystyle \implies \ \$ $\displaystyle F$ $=$ $\displaystyle \int \frac {g \left({u}\right)} {f' \left({x}\right)} \mathrm d u$ Definition of Primitive

$\blacksquare$