# Primitive of Composite Function

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## Theorem

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

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

Then:

- $\displaystyle \int \map {g \circ f} x \rd x = \int \frac {\map g u} {\map {f'} x} \rd u$

where $u = \map f x$.

## Proof

\(\displaystyle \map F x\) | \(=\) | \(\displaystyle \int \map {g \circ f} x \rd x\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int \map g {\map f x} \rd x\) | Definition of Composition of Mappings | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \int \map g u \rd x\) | where $u = \map f x$ | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d F} {\d x}\) | \(=\) | \(\displaystyle \map g u\) | Definition of Primitive (Calculus) | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d F} {\d u} \frac {\d u} {\d x}\) | \(=\) | \(\displaystyle \map g u\) | Chain Rule for Derivatives | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d F} {\d u} \frac \d {\d x} \map f x\) | \(=\) | \(\displaystyle \map g u\) | Definition of $u$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d F} {\d u} \map {f'} x\) | \(=\) | \(\displaystyle \map g u\) | Definition of Derivative | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \frac {\d F} {\d u}\) | \(=\) | \(\displaystyle \frac {\map g u} {\map {f'} x}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle F\) | \(=\) | \(\displaystyle \int \frac {\map g u} {\map {f'} x} \rd u\) | Definition of Primitive (Calculus) |

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 14$: General Rules of Integration: $14.6$