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zgl
United States1055 Posts
Indefinite integral of f'(x) / f(x) dx The answer is supposed to be log(f(x)) ? I tried integration by parts with u = 1 / f(x) du = - f'(x) / (f(x)^2) dx dv = f'(x) dx v = f(x) uv - integral( v du ) = f(x)/f(x) - integral( - f'(x)/f(x) dx ) = 1 + integral ( f'(x)/f(x) dx ) This gives 0 = 1 wtf... edit: can someone also explain where is the mistake that caused 0 = 1 ?? edit 2 : found solution yay <3 wikipedia d/dx ( ln ( f(x) ) = f'(x) / f(x) and take integral of both sides  | ||
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xmShake
United States1100 Posts
Derivative of ln(f(x)) = f '(x) / f (x) Integral of f '(x) / f (x) = ln | f(x) | | ||
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zgl
United States1055 Posts
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Cloud
Sexico5880 Posts
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gamecrazy
United States421 Posts
That would make dv = ln f(x) u = f'(x) du = f''(x) f'(x) ln f(x) - int( f ''(x) / f(x) ) Yeah i don't know what i'm doing either.. XD edit: Lol y didn't i think of that.... d/dx ln( f(x)) = f'(x)/ f(x) that was obvious  | ||
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toopham
United States551 Posts
its call the chainrule. | ||
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zgl
United States1055 Posts
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gamecrazy
United States421 Posts
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toopham
United States551 Posts
On July 05 2010 11:17 zgl wrote: Yeah. I just realized how silly it was to try to integrate directly when the answer is given. It would be nice to see how to actually integrate it though... You look at it and then you say ohh the answer is log(f(x)) Or you can do the long way. u = f(x) du = f'(x) dx U-substitution you get integration of (1/u)du Thus the answer is log(u) back substitute ---> log(f(x)) | ||
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Roggles
United States38 Posts
then it becomes integral of du / u, which is ln(u) + constant put u = f(x) back in and there you go standard u sub ^_^ | ||
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Cloud
Sexico5880 Posts
For example, you can find that ln(x^r) = rln(x) because when you derive either side of the equation you get the same result (r/x), with the left side you apply the chain rule, with the right side, you simply take the r as constant and leave it out of the derivation. | ||
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n.DieJokes
United States3443 Posts
du=f'(x)dx so int. (f'(x)/u)(du/f'(x)) f'(x) cancels, you left with int. (1/u)du ln(u)+c An: ln(f(x))+c Hmm, I guess I did it wrong too. Sorry buddy Edit: Can someone tell me why I'm wrong. Edit 2: Are you sure it's not ln? Cause then I'm right. I forgot how ocd I am about math >.> | ||
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zgl
United States1055 Posts
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