Integration by Parts

Theorem
Let $f$ and $g$ be real functions which are continuous on the closed interval $\left[{a \,.\,.\, b}\right]$.

Let $f$ and $g$ have primitives $F$ and $G$ respectively on $\left[{a \,.\,.\, b}\right]$.

Then:
 * $\displaystyle \int_a^b f \left({t}\right) G \left({t}\right) \rd t = \left[{F \left({t}\right) G \left({t}\right)}\right]_a^b - \int_a^b F \left({t}\right) g \left({t}\right) \rd t$

This is frequently written as:
 * $\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \ \frac {\d u} {\d x} \rd x$

or:
 * $\displaystyle \int u \rd v = u v - \int v \rd u$

where it is understood that $u, v$ are functions of the independent variable.

Proof
By Product Rule for Derivatives:
 * $D \left({FG}\right) = f G + F g$

Thus $F G$ is a primitive of $f G + F g$ on $\left[{a \,.\,.\, b}\right]$.

Hence, by the Fundamental Theorem of Calculus:
 * $\displaystyle \int_a^b \left({f \left({t}\right) G \left({t}\right) + F \left({t}\right) g \left({t}\right)}\right) \rd t = \left[{F \left({t}\right) G \left({t}\right)}\right]_a^b$

The result follows.