Integral of 1/x = ln(x) Problem has missing absolute value?

In summary, the given problem involves finding the integral of 1/x using calculus. However, there is a confusion regarding the use of absolute value signs in the argument of the natural log function. The solution is that since it is an initial value problem, the argument in the natural log is always positive, making the absolute value signs unnecessary.
  • #1
Jonnyb42
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Integral of 1/x = ln(x)... Problem has missing absolute value!?

This is an ODE problem and solution, but what I'm not understanding here is calculus based:

2rwatya.jpg


Why does the argument to the red-boxed ln() not in absolute value signs?? ( | | )
u = cos(t) so the numerator can be negative. I am just confused as to why they left that out.

Thanks for any help.
 
Last edited:
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  • #2


My solution is that, it is an Initial value problem, (there is only one t which we care about) and with that t the argument in the natural log is positive and so u can ignore the abs value!
 

1. What is the integral of 1/x?

The integral of 1/x is ln(x), where ln(x) represents the natural logarithm of x.

2. Why does the problem mention a missing absolute value?

The problem likely mentions a missing absolute value because the integral of 1/x can have different solutions depending on whether the x value is positive or negative. The absolute value ensures that the correct solution is chosen regardless of the sign of x.

3. How is the integral of 1/x related to the natural logarithm?

The integral of 1/x is equal to the natural logarithm of x, which is represented by ln(x). This relationship can be derived through the process of integration by parts.

4. Can the integral of 1/x be simplified further?

Not really. The integral of 1/x is already in its most simplified form as ln(x). However, it is possible to use different notations, such as log(x), to represent the natural logarithm.

5. What is the domain of the integral of 1/x?

The domain of the integral of 1/x is all real numbers except for x = 0, as the natural logarithm is undefined at this point. In other words, the domain is (-∞, 0) ∪ (0, ∞).

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