Why Transform Integrals of Differential Functions?

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SUMMARY

The discussion centers on the transformation of integrals involving differential functions, particularly focusing on the integral of the natural logarithm, ln(x). Participants clarify that while the integral of ln(x) can be expressed in terms of a differential transformation (Integral(ln(x) * 1/x dx)), this is not universally applicable to all functions, such as algebraic ones. The correct method for solving the integral of ln(2x) is confirmed to involve integration by parts, leading to the result x*ln(2x) - x. Misunderstandings regarding the application of u-substitution and the differentiation of integrals are also addressed.

PREREQUISITES
  • Understanding of integral calculus, specifically integration by parts.
  • Familiarity with transcendental functions and their properties.
  • Knowledge of u-substitution in integral calculus.
  • Basic differentiation techniques and their application to integrals.
NEXT STEPS
  • Study integration by parts in detail, focusing on its application to logarithmic functions.
  • Learn about u-substitution and its role in simplifying integrals.
  • Explore the properties of transcendental functions versus algebraic functions in calculus.
  • Practice solving integrals involving natural logarithms and their transformations.
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Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of integral transformations and logarithmic functions.

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

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?
 
Dumbledore said:
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?



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



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?
 

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