# Differentials within integrals

1. Dec 6, 2008

### Dumbledore

Hello.

Can someone please explain why I have to transform an integral of a differential function into the form Integral ( lnx 1/x dx ) for example, for Integral ( lnx ).

It seems to only be done with transcendental functions and not the algebraic ones... ie. Integral ( x^2 ) != Integral ( x^2 2x dx)

Whereas, Integral (ln x) == Integral ( ln x 1/x dx)

2. Dec 6, 2008

### aerospaceut10

Well, the idea of an integral is that there will always be a requirement for the differential operator as it is essentially a summation of infinitely small width rectangular areas.

Taking the integral of (Ln(x)) is basically just Integral ( Ln(x) dx) and to solve that one would use integration by parts. I'm not sure where you're getting that Integral (ln x) == Integral ( ln x 1/x dx) term.

3. Dec 6, 2008

### Dumbledore

I think I get it, its because the differential of x is dx, and the differential of u is du. So if you have a function of a function you have to identify u and find du.

In ln x
u = x
du = dx

So in this case I am incorrect to say Integral (ln x) == Integral ( ln x 1/x dx)

But if it were ln (2x) then it would be Integral (ln (2x)) == Integral (ln(2x) 2dx)

Correct?

4. Dec 6, 2008

### aerospaceut10

Mmm...well.

The integral of ln (2x) is simply just Integral (Ln(2x) dx). Simple as that.

And then for actually solving this integral you would need to integrate by parts.

Are you trying to apply the u substitution with your statements of u = x or something?

5. Dec 7, 2008

### Dumbledore

I'm pretty sure that is incorrect, but I don't have the math background to know for sure. I see that I did make yet another mistake though... I'll show you how you can solve this without integrating by parts:

Integral( ln(2x) ) = 1/2 Integral ( ln(2x) 2dx) = 1/2 (1/2x) (2) = 1/2x

Is this not correct?

6. Dec 7, 2008

### aerospaceut10

Sorry, but that is not correct. You need to do integration by parts.

The integral of (ln(2x)) = x*ln(2x) - x . You can differentiate it again to see that it equals ln(2x).

If you differentiate 1/(2x) that does not get you ln(2x), it just goes to some x^-2 term.

7. Dec 7, 2008

### Dumbledore

Yeah you are right. Basically, everything I said is completely incorrect. This entire thread is an embarrassment.

8. Dec 7, 2008

### aerospaceut10

Nah, it's all good, that's why we have these forums! Do you still have any sort of misunderstandings or confusions about this particular question?