Answer
$${\text{The product rule}}$$
Work Step by Step
$$\eqalign{
& {\text{The integration by parts is based on the product rule of derivatives}}{\text{.}} \cr
& {\text{Demostration:}} \cr
& {\text{Let the functions }}u\left( x \right){\text{ and }}v\left( x \right){\text{ then,}} \cr
& \frac{d}{{dx}}\left[ {u\left( x \right)v\left( x \right)} \right] = u\left( x \right)v'\left( x \right) + v\left( x \right)u'\left( x \right) \cr
& d\left[ {u\left( x \right)v\left( x \right)} \right] = \left[ {u\left( x \right)v'\left( x \right) + v\left( x \right)u'\left( x \right)} \right]dx \cr
& {\text{By integrating both sides}},{\text{ we can write this rule in terms of an }} \cr
& {\text{indefinite integral}}: \cr
& u\left( x \right)v\left( x \right) = \int {\left[ {u\left( x \right)v'\left( x \right) + v\left( x \right)u'\left( x \right)} \right]} dx \cr
& {\text{Rearranging this expression in the form}} \cr
& \int {u\left( x \right)} \underbrace {v'\left( x \right)dx}_{dv} = u\left( x \right)v\left( x \right) - \int {v\left( x \right)\underbrace {u'\left( x \right)dx}_{du}} \cr
& \int u dv = uv - \int {vdu} \cr} $$
