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

1. ## 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!

2. Pull out a 1? There is a 1 in every equation since multiplying by 1 does nothing.

Just doing integration by parts will work.

3. 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?

4. 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.

5. 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.

6. Why pull out something that already exists?

7. Is this what you're doing? If it's, then it's valid, of course:

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

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

$\displaystyle = x\ln{x}-x+k = x\left(\ln{x}-1\right)+k.$

8. Yeah, that's what I'm doing. Thanks.