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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 :smile:
Reply 1
Avatar for f1mad
f1mad
13
ln(x)= ln(x)*1

Use IBP: let u= ln(x), dv/dx=1.
Reply 2
Avatar for elldeegee
elldeegee
19
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 :smile:



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,
∫ln(x) dx=∫1×ln(x) dx \displaystyle \int ln(x)\ dx = \displaystyle \int 1 \times ln(x)\ dx
I can see that you know:
∫uv dx \displaystyle \int uv\ dx
=[uv]−∫vdudx dx = \left[ uv \right] - \displaystyle\int v \frac{du}{dx}\ dx

and then let u u equal the one that can be differentiated (ln(x)ln(x)) and v v be the other (11)

Back to your actual question :

dudx=1\frac{du}{dx} =1 and what has happened is that they've multiplied up by "dx" so that they have got du =dx.
(edited 12 years ago)
Reply 3
Avatar for raheem94
raheem94
13
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 :smile:


See this example:

Thanks guys I understand now )

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