why does integration by parts work?

Question

$$\int u(x)v'(x)\;\text dx = u(x)v(x) - \int u'(x)v(x)\;\text dx$$ or more compactly: $$\int u\;\text dv = uv - \int v\;\text du$$ I get how we get the formula, my main question is why $$\int u \;dv$$

is considered the integral of $uv$; it looks like it's the integral of $u$ with respect to $\text d v$ which isn't the same?

Answer 1

$\int uv’dx=\int udv$

You can see this using abuse of notation:

$\int uv’dx=\int u \frac{dv}{dx}dx=\int udv$

I call this an abuse because in reality, this being true is why we use the notation that we use, not the other way around. Still, it should make sense from the chain rule.

Now, the rest of integration by parts is just the product rule for differentiation, but “backwards.” That should be clear just from rearranging some terms around.

For completeness, I will show it here:

$\frac{d}{dx}(uv)=uv’+vu’$

That is simply the product rule. Now subtract $vu’$ from each side and then integrate with respect to $x$, resulting in:

$\int d(uv)-\int vdu=\int udv$

$uv-\int vdu=\int udv$

Which was what we aimed to show.

Answer 2

First take the power rule. $$\frac{d}{dx}[uv] = u'v + uv'$$ Now $$u'v = \frac{d}{dx}[uv] - uv'$$ Now integrate both sides. $$\int{u'v} = uv - \int{uv'}$$

Answer 3

Product rule: $$\frac{d}{dx}(f(x)g(x)) = f'(x)g(x)+g'(x)f(x)$$ Integrate both sides $$f(x)g(x) = \int f'(x)g(x) \thinspace dx+\int g'(x)f(x) \thinspace dx$$

$$\int f'(x)g(x) \thinspace dx=f(x)g(x)-\int g'(x)f(x) \thinspace dx$$ Let $f(x) = v, \thinspace g(x) = u$ $$\int u \thinspace dv=uv- \int v \thinspace du$$