integrating ln(X)
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integrating ln(X)So I tyed in "integrate ln(x)" into wolphram alpha and I get
xln(x)-x
now the steps it gives are doing it by parts obviously
U=ln(x) dV= dx
dU = 1/x V= X
since when can we include the dx at the end of every intergral with respect to x, as part of our integration ?
can someone explain
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Re: integrating ln(X)(Original post by kingkongjaffa)
So I tyed in "integrate ln(x)" into wolphram alpha and I get
xln(x)-x
now the steps it gives are doing it by parts obviously
U=ln(x) dV= dx
dU = 1/x V= X
since when can we include the dx at the end of every intergral with respect to x, as part of our integration ?
can someone explain
I think you've assumed that because dv = dx, they have just taken the end "dx" and used that as v.
in fact they have used the fact that f1mad has stated,
I can see that you know:
![= \left[ uv \right] - \displaystyle\int v \frac{du}{dx}\ dx](http://www.thestudentroom.co.uk/latexrender/pictures/4e/4eedae4bbc41522719e414d5f4f53576.png "= \left[ uv \right] - \displaystyle\int v \frac{du}{dx}\ dx")
and then let
equal the one that can be differentiated (
) and 
be the other (
)
Back to your actual question :
and what has happened is that they've multiplied up by "dx" so that they have got du =dx.
Last edited by elldeegee; 18-04-2012 at 03:13. -
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Re: integrating ln(X)See this example:(Original post by kingkongjaffa)
So I tyed in "integrate ln(x)" into wolphram alpha and I get
xln(x)-x
now the steps it gives are doing it by parts obviously
U=ln(x) dV= dx
dU = 1/x V= X
since when can we include the dx at the end of every intergral with respect to x, as part of our integration ?
can someone explain
