Question

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

Answer 1

easiest way is probably\[u=t+1\implies t=u-1\]\[du=dt\]

Answer 2

long division would also work

Answer 3

I tried it with long division and it was pretty easy that way.

Answer 4

i like the gimmick of writing
\[3t-2=3t+3-5\]

Answer 5

where do I go with long division after I get 3 remainder -5?

Answer 6

so you get
\[\frac{3t-2}{t+1}=\frac{3t+3-5}{t+1}=1-\frac{5}{t-1}\]

Answer 7

oops
\[\frac{3t-2}{t+1}=\frac{3t+3-5}{t+1}=3-\frac{5}{t-1}\]

Answer 8

then integrate each piece
first one gives you \(3x\) second gives you \(-5\ln(t-1)\)

Answer 9

ah got it thanks very much

Answer 10

damn another typo!!
should be
\[3x-5\ln(x+1)\]