Confused about integral of f'(x)/f(x)

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SUMMARY

The discussion centers on the confusion surrounding the integration of the functions f'(x)/f(x) and the implications of constants in logarithmic functions. Specifically, the integrals of the two equivalent fractions, 2x/x² and 6x/3x², yield ln(x²) and ln(3x²) respectively. The difference arises from the constant factor, where ln(3x²) can be expressed as ln(3) + ln(x²), highlighting the importance of the constant of integration (+C) in determining the final result. This illustrates that while the integrands are equivalent, their integrals differ by a constant term.

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  • Familiarity with logarithmic properties, specifically ln(ab) = ln(a) + ln(b).
  • Knowledge of algebraic manipulation of fractions.
  • Experience with the concept of the constant of integration (+C).
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Lucy Yeats
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A fraction obviously stays the same if you multiply top and bottom by the same constant. If you integrate two equal functions, surely you should get the same answer? By applying the rule in the title, you can get different answers. I'm very confused.

Example: 2x/x^2=6x/3x^2
The integral of 2x/x^2 is ln(x^2).
The integral of 6x/3x^2 is ln(3x^2)

How come these aren't equal? Is it to do with the +c?
 
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Remember that \ln (ab) = \ln a + \ln b. Hence \ln(3x^2)=\ln(x^2)+\ln(3), so the two expressions just differ by a constant.
 
Ah, okay. Thanks!
 
Lucy Yeats said:
A fraction obviously stays the same if you multiply top and bottom by the same constant. If you integrate two equal functions, surely you should get the same answer? By applying the rule in the title, you can get different answers. I'm very confused.

Example: 2x/x^2=6x/3x^2
The integral of 2x/x^2 is ln(x^2).
The integral of 6x/3x^2 is ln(3x^2)

How come these aren't equal? Is it to do with the +c?

Yes, it is about the plus C.
ln(3x^2)=ln(3)+ln(x^2)
 

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