Calculus: Early Transcendentals, Chapter 7, 7.4, Section 7.4, Problem 8

int ((3t-2)/(t+1)) dt
sol:
let t+1 = u
=> t= u-1
=> dt =du
int ((3t-2)/(t+1)) dt
= int ((3(u-1)-2)/(u)) du
= int ((3u-3-2)/(u)) du
= int (3u-5/u) du
= int (3- (5/u)) du
= int 3 du - int (5/u) du
= 3u - 5*ln(u)
= 3*(t+1)- 5*ln(t+1) +C
:)