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integral 1/cy-1 dy
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\[\int\limits \frac{1}{cy-1}dy\]
with c being a constant
i would think the answer is: ln(cy-1) , but it's not
+C
still wrong
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answer should be: 1/c . ln(cy-1)
for some reason..
\[\int\limits \frac1{cy-1}\,\mathrm dy\\=\frac1c\int\frac1{y-1/c}\,\mathrm dy\\ =\frac1c\log{(y-1/c)}+C\]
no, don't think that's right checked wolframalpha
nevermind, figured it out
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used substitution
\[\int\limits \frac1{cy-1}\,\mathrm dy\\=\frac1c\int\frac c{cy-1}\,\mathrm dy\\ =\frac1c\log{(cy-1)}+C\] its the same thing
what did you get ?
the same
used u= cy-1, etc..
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got 1/c * ln(cy-1)
thanks
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