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

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

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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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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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