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 $$\int_a^b f \left({t}\right) G \left({t}\right) dt = \left[{F \left({t}\right) G \left({t}\right)}\right]_a^b - \int_a^b F \left({t}\right) g \left({t}\right) dt$$.

This is frequently written as:
 * $$\int u \, dv = u v - \int v \, du$$

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

Proof
By Product Rule for Derivatives, we have $$D \left({FG}\right) = f G + F g$$.

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

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

The result follows.