Solving Integral from Table: Differences Explained

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SUMMARY

The integral of the function \(\int \frac{1}{a^2-u^2}du\) is confirmed to yield the result \(\frac{1}{2a} \ln\left|\frac{u+a}{u-a}\right| + c\) as stated in two different mathematical texts. The discussion highlights a derivation that leads to \(\frac{1}{2a} \ln\left|\frac{a+u}{a-u}\right| + c\), which is equivalent due to the properties of logarithms and absolute values. The interchangeability of \(u+a\) and \(a+u\) is clarified, while the difference between \(u-a\) and \(a-u\) is attributed solely to the absolute value property, confirming that \(|a-u| = |u-a|\).

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In a table in two different books they both say:
[tex]\int \frac{1}{a^2-u^2}du=\frac{1}{2a} ln\left|\frac{u+a}{u-a}\right|+c[/tex]

but I am not having the same result:

[tex]\int \frac{1}{a^2-u^2}du=\frac{1}{2a} \left(\int \frac{1}{a-u}du + \int \frac{1}{a+u}du\right) = \frac{1}{2a} \left(-ln\left| a-u \right|\right)+ln\left| a+u \right|+c = \frac{1}{2a} ln\left|\frac{a+u}{a-u}\right|+c[/tex]


Obviously the u+a at top is interchangeable but I am not seeing an explanation for the bottom (u-a vs. a-u) being different. Is it just an absolute value thing?
 
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Well |a-u| = |u-a|, so yes I would say its just an absolute value thing.
 

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