I need to find the integral: integral (t-1)/(t+1) dt

Note that t - 1 = t + 1 - 2.

∫(t - 1)/(t + 1) dt = ∫(t + 1 - 2)/(t + 1) dt = ∫1 - 2/(t + 1) dt

= t - 2*ln|t + 1| + C

∫[(t-1)/(t+1)]dt

t + 1 divides into t - 1 with a quotient of 1 and a remainder of -2 as follows:
1
.......┌────
t + 1│t - 1
........-t - 1
........────
.............-2


Therefore, (t - 1)/(t + 1) becomes 1 - 2/(t + 1) and the integral becomes two integrals:

∫[1]dt -2∫[1/(t+1)]dt = t - 2ln|t +1| + c

Divide to obtain 1 - [ 2 / (t + 1) ]

I = ∫ 1 - 2/(t + 1) dt

I = t - 2 log (t + 1) + C

(t-1)/(t+1) dt =
1 - 2/(t+1) dt

t - 2ln(t+1) + C

t - 2 Log[1 + t] + C