Question

Question #72bda

Answer 1

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`