When deriving the integration by parts formula, you can use the product rule to do so,
i.e. $\{uv\}' = uv' + vu'$
$\Rightarrow \int \{uv\}' = \int udv + \int vdu$
hence $uv = \int udv + \int vdu$.
If $uv$ is the integral of $\{uv\}'$ then why is the formula rearranged to read
$$\int udv = uv-\int vdu ?$$
I am confused as to why this is as it seems you are finding half of the product rule by subtracting the other half of the product rule from the integrated function, it just seems counter intuitive.
Best Answer
$\text{ Say you want to evaluate } \\ I=\int \ln(x) dx \\ \text{ we will first write } \\ I \text{ as an integral of a product so we can apply the integrate by parts formula }\\ I=\int 1 \cdot \ln(x) dx \\ \text{ so this is in the form } 1 \cdot \ln(x) dx \\ \text{ is in the form } u \cdot dv \\ \text{ we just need to choose which is which and we want to do so so it benefits us in the end }\\ \text{ I will choose } u=\ln(x) \text{ and } dv=1 dx \\ \text{ so as you said the formula is given as } \\ \int u dv=uv-\int v du \\ \text{ so if } u=\ln(x) \text{ then } \frac{du}{dx}=\frac{1}{x} \text{ or } du=\frac{1}{x} dx \text{ and if } dv=1 dx \text{ then } v=x \\ \text{ so we can use the formula } \\ \int u dv=uv-\int v du \text{ to rewrite our integral } I \text{ as } \ln(x) \cdot x - \int x \cdot \frac{1}{x} dx \\ \\ \text{ so we see that our integral is actually doable now with the help of this formula } \\ \text{ so it isn't counterproductive to write the product rule in this way }$