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integral of (3t-2)/(t+1)

Mathematics
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easiest way is probably\[u=t+1\implies t=u-1\]\[du=dt\]
long division would also work
I tried it with long division and it was pretty easy that way.

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Other answers:

i like the gimmick of writing \[3t-2=3t+3-5\]
where do I go with long division after I get 3 remainder -5?
so you get \[\frac{3t-2}{t+1}=\frac{3t+3-5}{t+1}=1-\frac{5}{t-1}\]
oops \[\frac{3t-2}{t+1}=\frac{3t+3-5}{t+1}=3-\frac{5}{t-1}\]
then integrate each piece first one gives you \(3x\) second gives you \(-5\ln(t-1)\)
ah got it thanks very much
damn another typo!! should be \[3x-5\ln(x+1)\]
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