4 ms·
So, your proposed solution to integral(ln(x)) involves first noticing that d/dx (1/x) = ln(x), then proceeding to use v=1/x to integrate by parts? I think I am
by gizmo686 10y ago
So, your proposed solution to integral(ln(x)) involves first noticing that d/dx (1/x) = ln(x), then proceeding to use v=1/x to integrate by parts?
I think I am misunderstanding something, because as far as I can tell, integration by parts where one of the parts is 1 is literally useless.
- yequalsx 10y agou = ln(x) and dv = 1 du = 1/x and v = x integral ln(x) = x ln(x) - integral 1/x times x integral ln(x) = x ln(x) - x + C I could give integral arctan(x) but with the advent of computer algebra systems I'm mostly interested in them knowing the basic examples and to not burden them on a test with something more complicated. EDIT: The derivative of 1/x is not ln(x) as you stated. You got it backwards and my guess is that is the source of your confusion.