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# integrating ln(X) Watch

1. 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
2. ln(x)= ln(x)*1

Use IBP: let u= ln(x), dv/dx=1.
3. (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:

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.
4. (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
See this example:

5. Thanks guys I understand now )

Updated: April 18, 2012
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