# Integration By Parts While the substitution rule can be considered the “inverse chain rule”, the technique of **integration by parts** fills the role of the “*inverse product rule*”. > [!summary] Integration by Parts > Let $u=f(x)$ and $v=g(x)$ be functions with continuous derivatives. Then, the integration by parts formula for the integral involving these two functions is: > $\int u dv = uv - \int v du$ ## Deriving The Integration By Parts Formula If $h(x)=f(x)g(x)$, then by the product rule of differentiation, we obtain: $h'(x)=f'(x)g(x)+g'(x)f(x)$ Although it may seem counterproductive, let’s now integrate both sides of this equation: $\begin{align} \int h'(x)dx &= \int [g(x)f'(x) + f(x)g'(x)]dx \\ &= \int g(x)f'(x) dx + \int f(x)g'(x) dx \end{align}$ Since $\int h'(x)dx = h(x)$, and $h(x) = f(x)g(x)$, this gives us $f(x)g(x) = \int g(x)f'(x)dx + \int f(x)g'(x) dx$ Solving for $\int f(x)g'(x) dx$, we obtain $\int f(x)g'(x) dx = f(x)g(x) - \int g(x)f'(x)dx$ By making the substitutions $u=f(x)$ and $v=(gx)$, which in turn make $du=f’(x)dx$ and $dv=g’(x)dx$, we can write this in the more compact form $\int u dv = uv - \int v du$ > [!attention] > For basic integration by parts problems, the key is to chosoe $u$ and $dv$ such that: > - $dv$ is easy to integrate. > - $du$ is simpler than $u$. > ## Performing Integration By Parts ### Variable Selection Using LIATE The first ### The DI/Tabular Method > [!NOTE] What’s the point of the DI method? > > Simplifying Multi- > > There are many situations where you’ll need to use integration by parts multiple times to reach a final solution.