Question

How do I solve the integral of (3t-2)/(t+1) dt ?

Answer 1

do long division..:)

Answer 2

Try \(u=t+1\)

Answer 3

\[
3t-2 = 3t - 3-2+3 = 3(t-1)+1 = 3u+1
\]

Answer 4

(3t-2)/(t+1) = A + B/(t + 1)
A (t + 1) + B = 3t - 2 : A = 3, B = -5
now it's easily integrated

Answer 5

i thought u = t+1, not t-1

Answer 6

@Yahoo! has it right... divide then integrate