# Is this a legitimate way of finding the integral of lnx?

• Nov 30th 2010, 05:59 PM
Glitch
Is this a legitimate way of finding the integral of lnx?
$\int{\ln{x} \ dx}$

= $\int{\ln{x}(1) \ dx}$ //pull out a '1'

= $\ln{x}(x) - \int{\frac{1}{x}x}$ //use integration by parts

= $x\ln{x} - \int{1}$

= $x\ln{x} - x$

= $x(\ln{x} - 1)$

Is pulling out a 1 in the first step a legitimate method when using integration by parts, or is it just a coincidence that it works? Thanks!
• Nov 30th 2010, 06:00 PM
dwsmith
Pull out a 1? There is a 1 in every equation since multiplying by 1 does nothing.

Just doing integration by parts will work.
• Nov 30th 2010, 06:04 PM
Glitch
I think my understanding of integration by parts isn't too good. :(

I'm not sure how to perform it without there being two functions to work with, which is why I pulled out the one. What's the proper way of dealing with this?
• Nov 30th 2010, 06:07 PM
dwsmith
If we have an x, there is always an implied 1. There is a one in every function you can think since multiplying by a 1 does nothing to an equation.
• Nov 30th 2010, 06:14 PM
Glitch
So my method was right, but you don't think of it as pulling out a one? I just did that to help my head get around it.
• Nov 30th 2010, 06:15 PM
dwsmith
Why pull out something that already exists?
• Nov 30th 2010, 06:23 PM
TheCoffeeMachine
Is this what you're doing? If it's, then it's valid, of course:

$\displaystyle \int\ln{x}\;{dx} = \int(\ln{x})({1})\;{dx} = \int(\ln{x})(x)'\;{dx}$

$\displaystyle = (\ln{x})(x)-\int\left(\ln{x}\right)'\cdot x\;{dx} = (\ln{x})(x)-\int (1)\;{dx}$

$\displaystyle = x\ln{x}-x+k = x\left(\ln{x}-1\right)+k.$
• Dec 1st 2010, 01:56 AM
Glitch
Yeah, that's what I'm doing. Thanks.