Question #72bda

1 Answer
Mar 2, 2017

Your answer is correct, it just has a different C.

Explanation:

(t+1/2)/(t+2/3) = (1/2(2t+1))/(1/3(3t+2)) = 3/2 (2t+1)/(3t+2)

So

log abs((t+1/2)/(t+2/3) ) +C_1 = log(3/2 (2t+1)/(3t+2)) +C_1

= log(3/2) + log abs( (2t+1)/(3t+2)) +C_1

= log abs( (2t+1)/(3t+2)) +C_2