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KSCphysics
Apr1-04, 04:42 PM
One of my recent problems is with double integrals... and im having a brain fart on the \int ln(u) du can you do this?

NateTG
Apr1-04, 04:48 PM
You can do it by parts. Consider that
\frac{d}{dx} x \ln x = \ln x + 1

KSCphysics
Apr1-04, 06:52 PM
hrmm... i see what your saying..

iluvsr20s
Apr1-04, 08:47 PM
wouldn't it just be 1/u???
unless that little dash is a negative sign mean your are integrating 1/ln(u) then I'm not sure but i think it would just be u then, but i am probably wrong

ahrkron
Apr1-04, 08:55 PM
iluvsr20s,

You are probably confused with the fact that \int \frac{1}{u} du = \ln x + c. It is not the other way around.

philosophking
Apr1-04, 10:18 PM
... hehe. this ain't too bad. Can we change it to int(ln(x),dx) though? It's how I've always done it notation-wise.

let u = ln(x), dv = dx --> du = 1/x(dx), v = x

So, u*v -int(v,du) = int(u,dv)

x*ln(x) - int(1,dx) = int(ln(x),dx)
x*ln(x) - x = int(ln(x),dx)

Et voila.