Proving alternative of Integration by Parts from Stegun's Book and its Definite form

I only knew 1 kind of integration by parts given by this: $$\int u\, \Bbb dv = uv - \int v \,\Bbb du\tag{1}$$ And i can understand where it came from.

And the definite integral form is: $$\int_a^b u\, \Bbb dv = uv\,\Bigg|_a^b - \int_a^b v \,\Bbb du\tag{2}$$

From Stegun's Book, i found another interesting form of integration by parts given by: $$\int uv \Bbb dx = \left(\int u\Bbb dx\right)v - \int\left(\int u \Bbb dx\right)\frac{\Bbb dv}{\Bbb dx} \Bbb dx\tag{3}$$.

Maybe i can derive it using the product rule of differentation:

\begin{align} D_x\left[(u(x)v(x))\right] &= u'(x)v(x) + u(x)v'(x) \\ \int D_x\left[(u(x)v(x))\right] \Bbb dx&= \int u'(x)v(x) + u(x)v'(x)\Bbb dx\\ uv &= \int v \Bbb du + \int u \Bbb dv \end{align}

But what now?

My question is, what's the "definite" integral form of Eq. $$(3)$$? And how to prove it? (I confuse with $$\frac{\Bbb dv}{\Bbb dx}$$) and which parts are evaluated from $$a$$ to $$b$$ for the definite integral form?

This is the same rule, it just uses different terms. In the second formula, $$u$$ is now $$v$$, and $$v$$ has become $$\int u\,dx$$, or put otherwise, $$dv$$ has become $$u\,dx$$.
• What about the definite integral form? Is it like $$\int_a^b uv \Bbb dx = \left(\int_a^b u\Bbb dx\right)v - \int_a^b\left(\int_a^b u \Bbb dx\right)\frac{\Bbb dv}{\Bbb dx} \Bbb dx$$. ? Jan 22 at 2:47
• Are $v$ and $v'$ evaluated from $a$ to $b$ or not? Jan 22 at 2:48