Integration by Parts with a Jump Discontinuity I ran into the following problem and its solution:

The integration by parts formula
  $$
\int_{a}^{b}u\frac{dv}{dx}\,dx=uv\bigg|_{a}^{b}-\int_{a}^{b}v\frac{du}{dx}\,dx
$$
  is known to be valid for functions $u(x)$ and $v(x)$, which are continuous and have continuous first derivatives. However, we will assume that $u$, $v$, $du/dx$, and $dv/dx$ are continuous only for $a\leqslant x\leqslant c$ and $c\leqslant x \leqslant b$; we assume that all quantities may have a jump discontinuity at $x=c$.
(a) Derive an expression for $\int_{a}^{b}u\,dv/dx\,dx$ in terms of $\int_{a}^{b}v\,du/dx\,dx$.
  $$
\int_{a}^{b}u\frac{dv}{dx}\,dx=uv\bigg|_{a}^{b}+uv\bigg|_{c^+}^{c^-}-\int_{a}^{b}v\frac{du}{dx}\,dx.
$$

Could anyone clarify to me how this was obtained?
Edit: Following the advice of Muphrid, I obtained the following:
$$\begin{align}
\int_{a}^{c^-}u\frac{dv}{dx}\,dx+\int_{c^+}^{b}u\frac{dv}{dx}\,dx&=uv\bigg|_{a}^{c^-}-\int_{a}^{c^-}v\frac{du}{dx}\,dx+uv\bigg|_{c^+}^{b}-\int_{c^+}^{b}v\frac{du}{dx}\,dx,\\
\int_{a}^{b}u\frac{dv}{dx}\,dx&=\color{red}{uv\bigg|_{a}^{c^-}+uv\bigg|_{c^+}^{b}}-\int_{a}^{b}v\frac{du}{dx}\,dx.
\end{align}$$
What is the rule for combining the terms in red?
 A: In response to your work from Muphrid:
$$\begin{align} 
\color{red}{uv\bigg|_{a}^{c^-}+uv\bigg|_{c^+}^{b}} &= uv(c^-) - uv(a) + uv(b) - uv(c^+) \\
&= uv(c^-) - uv(c^+) + uv(b) - uv(a) \\
&= uv\bigg|_{c^+}^{c^-} +uv\bigg|^b_a
\end{align}$$
A: I suggest using the usual integration by parts formula on the separate intervals $[a,c)$ and $(c, b]$ and then putting the two results together.
A: Corrected based on comment.
Consider the example
$$ \int_{0}^2 x \, H(x-1) \; dx, $$
where $H(x)$ is the unit step function. Modifying the limits to account for $H(x-1)$ gives
$$ \int_{0}^2 x \, H(x-1) \; dx= \int_{1}^2 x \; dx = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} $$
Applying standard integration by parts with $v = \frac{x^2}{2}$ and $u = H(x-1)$ gives the correct answer
$$ \int_{0}^2 x \, H(x-1) \; dx= \frac{x^2}{2} H(x-1) \bigg|_{0}^2 - \int_{0}^2 \frac{x^2}{2} \delta(x-1) \; dx = \frac{4}{2} - 0 - \frac{1}{2} = \frac{3}{2}. $$
Since $u$ has a jump at $c=1$, the modified result proposed in this question would not capture the $-\frac{1}{2}$ from the delta function and instead get the same value from
$$ \frac{x^2}{2} H(x-1) \bigg|_{1^+}^{1^-} = -\frac{1}{2}. $$
