What is integration by parts, and how is it useful?

"Integration by parts" is one of several methods in our arsenal that we can use to integrate a function f(x). It involves expressing f(x) as the product of two other functions, which I will call u(x) and v'(x) (where v' = ^dv/_dx is the first derivative of v with respect to x):
 
∫ f(x) dx = ∫ u(x) · v'(x) dx =  u(x) · v(x) - ∫ u'(x) · v(x) dx
 
This formula comes from the product rule for differentiation. If we differentiate the product u(x) · v(x) with respect to x using the product rule, we get the following:
 
^d/_dx(u(x) · v(x)) = u(x) · v'(x) + u'(x) · v(x)
 
Integrating both sides of this expression with respect to x removes the ^d/_dx on the left hand side, and creates two integrals on the right hand side:
 
u(x) · v(x) = ∫ u(x) · v'(x) dx + ∫ u'(x) · v(x) dx
 
If we subtract the final term from both sides, we arrive at our original formula again:
 
∫ f(x) dx = ∫ u(x) · v'(x) dx =  u(x) · v(x) - ∫ u'(x) · v(x) dx
 
But why would we want to use this method? Well, integration by parts is generally useful in cases such as the one below:
 
∫ Ln(x)/x² dx
 
At a first glance, we have no idea how to integrate this function. We do, however, know how to integrate 1/x² (which is our v' in this case), and we know how to differentiate Ln(x) (which is our u). So, using integration by parts:
 
∫ Ln(x)/x² dx = ∫ Ln(x) · (1/x²) dx = (-1/x) · Ln(x) - ∫ (-1/x) · (1/x) dx 
= -Ln(x)/x + ∫ (1/x²) dx = -Ln(x)/x - 1/x + C
 
(Not forgetting our constant of integration at the end!)