What is 1/(2a)(ln| u/a+1|-ln| u/a-1|)+C ?

The solution to 1/(2a)(ln| u/a+1|-ln| u/a-1|)+C is (ln(\frac{|u+a|)/(|u-a|))}{2a}+C

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simplify 1/(2a)(ln| u/a+1|-ln| u/a-1|)+C

simplify

\frac{1}{2 a} (ln \frac{u}{a} + 1 - ln \frac{u}{a} - 1) + C

\frac{ln ( \frac{∣ u + a ∣}{∣ u - a ∣} )}{2 a} + C

Solution steps

\frac{1}{2 a} (ln \frac{u}{a} + 1 - ln \frac{u}{a} - 1) + C

Apply log rule:

lo g_{c} (a) - lo g_{c} (b) = lo g_{c} (\frac{a}{b})

ln \frac{u}{a} + 1 - ln \frac{u}{a} - 1 = ln (\frac{\frac{u}{a} + 1}{\frac{u}{a} - 1})

= \frac{1}{2 a} ln (\frac{\frac{u}{a} + 1}{\frac{u}{a} - 1}) + C

Simplify

\frac{\frac{u}{a} + 1}{\frac{u}{a} - 1} : \frac{∣ u + a ∣}{∣ u - a ∣}

= \frac{1}{2 a} ln (\frac{∣ u + a ∣}{∣ u - a ∣}) + C

\frac{1}{2 a} ln (\frac{∣ u + a ∣}{∣ u - a ∣}) = \frac{ln ( \frac{∣ u + a ∣}{∣ u - a ∣} )}{2 a}

= \frac{ln ( \frac{∣ u + a ∣}{∣ u - a ∣} )}{2 a} + C

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What is 1/(2a)(ln| u/a+1|-ln| u/a-1|)+C ?
The solution to 1/(2a)(ln| u/a+1|-ln| u/a-1|)+C is (ln(\frac{|u+a|)/(|u-a|))}{2a}+C