Deriving the integration by parts formula 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. 
 A: $\text{ Say you want to evaluate } \\ I=\int \ln(x) dx \\ \text{ we will first write } \\  I \text{ as an integral of a product so we can apply the integrate by parts formula }\\ I=\int 1 \cdot \ln(x) dx \\ \text{ so this is in the form } 1 \cdot \ln(x) dx \\ \text{ is in the form } u \cdot dv \\ \text{ we just need to choose which is which and we want to do so so it benefits us in the end }\\ \text{ I will choose } u=\ln(x) \text{ and } dv=1 dx \\ \text{ so as you said the formula is given as } \\ \int u dv=uv-\int v du \\ \text{ so if } u=\ln(x) \text{ then } \frac{du}{dx}=\frac{1}{x}  \text{ or } du=\frac{1}{x} dx \text{ and if } dv=1 dx \text{ then } v=x \\  \text{ so we can use the formula } \\ \int u dv=uv-\int v du \text{ to rewrite our integral } I \text{ as } \ln(x) \cdot x - \int x \cdot \frac{1}{x} dx \\ \\ \text{ so we see that our integral is actually doable now with the help of this formula } \\ \text{ so it isn't counterproductive to write the product rule in this way }$
A: The main reason the integration by parts formula is written that way is because it makes it clear that if you can compute the antiderivative $v$ of $dv$, and if $du$ becomes "simpler", e.g. if $u$ is a polynomial, then you can typically reduce the overall "complexity" of the integral by using integration by parts, and make progress toward a solution. For example, you can integrate $\int x^n e^x$ this way, for any positive integer $n$, by putting $dv = e^x$ and $u = x^n$, and repeating with $dv = e^x$ and $u$ equal to a polynomial of decreasing degree.
A: $(uv)'= u\mathrm{d}v + v\mathrm{d}u$ which on rearrangement becomes $u\mathrm{d}v = (uv)' - v\mathrm{d}u$. Integrating this equation we get the formula for 'integration by parts', obviously. Try it.
