Question regarding integrating (int(1/u))

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In summary, there is confusion regarding the integral \int\frac{1}{u} and why the answer sometimes includes absolute value symbols and sometimes does not. It is likely that the absolute value symbols are only necessary if a negative value of x could result in a negative value for u when substituted back in. However, it is not necessary to worry about this as the correct answer should always be ln(|u|). Writing ln(u) may work in certain cases but using absolute value symbols is the more accurate approach.
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Linday12
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Homework Statement


I'm confused about integrating something like [tex]\int[/tex][tex]\frac{1}{u}[/tex]
Sometimes the answer seems to be ln|u| and sometimes just ln(u), and I wasn't sure why it is different from problem to problem. (after subbing u back in; I'm using the u-sub method)

It looks like the answer should be very straight forward as I can't find an explanation, so I'm not seeing something. My only guess would be that the absolute value symbols are used only if a negative value of x could give a negative value for u when subbing back in, making the ln(blah) undefined (for my course anyways). Is this about right? Thanks.
 
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Don't worry about it. The integral should be ln(|u|) because deriving that gives you 1/u for both u>0 and u<0. Sometimes, writing ln(u) works because (1) it's convenient and (2) it often works out in the end even if u is negative and the absolute value sign is ignored.
 

FAQ: Question regarding integrating (int(1/u))

What is the process for integrating 1/u?

Integrating 1/u follows the same general process as integrating any other expression. First, use the power rule to rewrite 1/u as u^-1. Then, add 1 to the exponent to get u^0. Finally, use the power rule again to integrate and add the constant of integration.

Can the integral of 1/u be simplified further?

Yes, the integral of 1/u can be simplified further by using the natural logarithm. After integrating, the expression becomes ln|u| + C, where C is the constant of integration.

Are there any special cases to consider when integrating 1/u?

Yes, when integrating 1/u, it is important to consider the domain of u. If u is negative, the natural logarithm will be undefined. In this case, the integral can be rewritten as -ln|u| + C to account for the negative sign.

Can the integral of 1/u be solved using substitution?

Yes, substitution can be used to solve the integral of 1/u. Let u be the variable inside the parentheses, and du be the differential. Then, substitute u and du into the integral and solve using the power rule and the constant of integration.

How can the integral of 1/u be visualized graphically?

The integral of 1/u can be visualized as the area under the curve of the function 1/u. This area will be infinite, as the function approaches infinity on both ends. Additionally, the domain must be restricted to positive values only, as the natural logarithm is undefined for negative values.

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