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integration of a logarithmic function? (1 Viewer)

pikto

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i haven't actually managed to find instructions on doing this, that is obviously why i am asking, but if anyone can set me in the right direction then that would be tops!
just a simple one, or a link to a theory/method.
 

Nobuo Uematsu

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pikto said:
i haven't actually managed to find instructions on doing this, that is obviously why i am asking, but if anyone can set me in the right direction then that would be tops!
just a simple one, or a link to a theory/method.
Use integration by parts.
let u=the log function and dv\dx=1
 

Antwan23q

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thats if it was Ln but how do u find the derivative of a log function?
 
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A primitive of lnx is x(lnx-1)
The derivative of a log function is 1/x, if you're using base 'e', and if not, use the change of base laws to put it in base 'e'
 

Antwan23q

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yrtherenonames said:
A primitive of lnx is x(lnx-1)
The derivative of a log function is 1/x, if you're using base 'e', and if not, use the change of base laws to put it in base 'e'
i dont follow
 

withoutaface

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Integrate by parts, like so:
int(lnx dx) = xlnx + int(x * 1/x dx) = xlnx + int(1)dx = xlnx + x + C
 

Nobuo Uematsu

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antwan2bu said:
yeh but how do you do
Int (logx dx)?
int. ln x dx
let u=lnx and dv\dx=1
du\dx=1\x and v=x
hence int. lnx dx=xlnx-int. x\x dx.......
 

Antwan23q

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I already know that! im not talk about a log to base e, im talkin about a log to base 10
 

Nobuo Uematsu

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antwan2bu said:
I already know that! im not talk about a log to base e, im talkin about a log to base 10
"Today 10:04 PM
yrtherenonames A primitive of lnx is x(lnx-1)
The derivative of a log function is 1/x, if you're using base 'e', and if not, use the change of base laws to put it in base 'e'
"
 
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Mate, you know that logx base b=lnx/lnb
Just log laws, so a log of something other than base e is just a constant 1/(lnb) times the log of that in base e

So integral of logx base 10=integral of lnx/ln10
=ln10(integral of lnx)
=x.ln10(lnx-1)
 

Antwan23q

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i was thinkin along the lines of ln x = log x / log e
and got messed up
 

haboozin

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antwan2bu said:
i was thinkin along the lines of ln x = log x / log e
and got messed up
wouldnt that be

lnx/lne
=
ln x


since log<sub>e</sub> e = 1 by definition.

(e to the power of what gives you e -- answer being 1)
 
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Yeh he was doing it backwards, using change of base to change lnx to logx instead of vice versa.
 

rama_v

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Log x = ln x ...thats the notation thats used (in NSW anyway). Log the the base 10 is just that, its log10x
 

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