# Integration by Parts

In differential calculus, we derived formulas to differentiate products of functions.

Recall that

$\frac{d\ \color{teal}f(x) \color{green}g(x)}{dx} = \color{teal}f'(x)\color{green}g(x)\color{currentcolor}+\color{teal}f(x)\color{green}g'(x)\color{currentcolor}.$

Integration by Parts is a method to do the same sort of thing, for integrals. The formula can be derived simply by re-arranging the product rule (shown above).

In order to see this a bit better, we'll perform the following substitution:

$\textcolor{orchid}{h(x)} = \textcolor{teal}{f'(x)}.$

The Fundamental Theorem of Calculus tells us that

$\int \textcolor{orchid}{h(x)}\ dx = \textcolor{teal}{f(x)}.$

Let's substitute! Replacing all occurrences of $\textcolor{teal}{f(x)},$ we get:

$\frac{d\ \int \textcolor{orchid}{h(x)}\ dx\ \cdot \textcolor{green}{g\left(x\right)}}{dx}=\textcolor{orchid}{h\left(x\right)}\textcolor{green}{g\left(x\right)}+\int \textcolor{orchid}{h(x)}\ dx\cdot \textcolor{green}{g'\left(x\right)}.$

Now, there's a big ugly $\frac{d}{dx}$ here - let's get rid of it! Remember - integration kills differentiation, and vice-versa!

So, we integrate both sides.

$\textcolor{green}{g\left(x\right)} \int \textcolor{orchid}{h(x)}\ dx=$
$\int \textcolor{orchid}{h(x)}~\textcolor{green}{g\left(x\right)}\ dx+ \int\textcolor{green}{g'\left(x\right)}\int \textcolor{orchid}{h(x)}dx \ dx.$

Rearranging to isolate $\int \textcolor{orchid}{h(x)}~\textcolor{green}{g\left(x\right)}\ dx$, we see that

$\int \textcolor{orchid}{h(x)}~\textcolor{green}{g\left(x\right)}\ dx =$
$\textcolor{green}{g\left(x\right)} \int \textcolor{orchid}{h(x)}\ dx - \int \textcolor{green}{g'\left(x\right)} \int \textcolor{orchid}{h(x)}\ dx \ dx.$

And, this is integration by parts!

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