Understanding Limits of Integration in Integration-by-Parts My understanding of integration-by-parts is a little shaky.  In particular, I'm not totally certain that I understand how to properly calculate the limits of integration.
For example, the formula I have is:
$\int_{v_1}^{v_2}{u dv} = (u_2 v_2 - u_1 v_1) - \int_{u_1}^{u_2}{v du}$
I'd like to see how to calculate $u_1$ and $u_2$, preferably in a complete example (that solves a definite integral.)  I'm really interested in an example where the limits of integration change; i.e. $u_1$ and $u_2$ are different than $v_1$ and $v_2$, if possible.
 A: Okay.
$$\int_1^2 \ln x \, dx = [x \ln x]_{x = 1}^2 - \int_1^2 1 \, dx = 2 \ln 2 - 1$$
A prototype example. Where $u = \ln x$ and $v = x$.
A: A more precise notation is this one
$$\int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx=\left(
u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right) -\int_{x_{1}}^{x_{2}}u^{\prime
}(x)v(x)dx$$
which is derived from the derivative rule for the product
$$(u(x)v(x))^{\prime }=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$$
or
$$u(x)v^{\prime }(x)=(u(x)v(x))^{\prime }-u^{\prime }(x)v(x).$$
So
$$\begin{eqnarray*}
\int_{x_{1}}^{x_{2}}u(x)v^{\prime }(x)dx
&=&\int_{x_{1}}^{x_{2}}(u(x)v(x))^{\prime }dx-\int_{x_{1}}^{x_{2}}u^{\prime
}(x)v(x)dx \\
&=&\left. (u(x)v(x))\right\vert
_{x=x_{1}}^{x=x_{2}}-\int_{x_{1}}^{x_{2}}u(x)v(x)dx \\
&=&\left( u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right)
-\int_{x_{1}}^{x_{2}}u^{\prime }(x)v(x)dx.
\end{eqnarray*}.$$
If you write $dv=v^{\prime }(x)dx$ and $du=u^{\prime }(x)dx$, you get your
formula but with $u,v$ as a function of $x$
$$\int_{v_{1}(x)}^{v_{2}(x)}u(x)dv=\left(
u(x_{2})v(x_{2})-u(x_{1})v(x_{2})\right)
-\int_{u_{1}(x)}^{u_{2}(x)}v(x)du$$
Example: Assume you want to evaluate $\int_{x_{1}}^{x_{2}}\log
xdx=\int_{x_{1}}^{x_{2}}1\cdot \log xdx$. You can choose $v^{\prime }(x)=1$
and $u(x)=\log x$. Then $v(x)=x$ (omitting the constant of integration) and 
$u^{\prime }(x)=\frac{1}{x}$. Hence
$$\begin{eqnarray*}
\int_{x_{1}}^{x_{2}}\log xdx &=&\int_{x_{1}}^{x_{2}}1\cdot \log xdx \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)
-\int_{x_{1}}^{x_{2}}\frac{1}{x}\cdot xdx \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)
-\int_{x_{1}}^{x_{2}}dx \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left(
x_{2}-x_{1}\right) 
\end{eqnarray*}$$

The same example with your formula:
$$u=\log x,v=x,dv=dx,v=x,du=\frac{1}{x}dx$$
$$u_{2}=\log x_{2},u_{1}=\log x_{1},v_{2}=x_{2},v_{1}=x_{1}$$
$$\begin{eqnarray*}
\int_{v_{1}}^{v_{2}}udv &=&\left( u_{2}v_{2}-u_{1}v_{2}\right)
-\int_{u_{1}}^{u_{2}}vdu \\
\int_{x_{1}}^{x_{2}}\log xdx &=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot
x_{1}\right) -\int_{\log x_{1}}^{\log x_{2}}xdu \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right)
-\int_{x_{1}}^{x_{2}}x\cdot \frac{1}{x}dx \\
&=&\left( \log x_{2}\cdot x_{2}-\log x_{1}\cdot x_{1}\right) -\left(
x_{2}-x_{1}\right). 
\end{eqnarray*}$$
Note: The limits of integration, although different in terms of $u(x),v(x)$, when expressed in terms of the same variable $x$ of functions $u(x),v(x)$ are the same in both sides.
For a strategy on how to chose the $u$ and $v$ terms see this question.
