Question #72bda

1 Answer
Jim H
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#

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