Integrals of Fcns Using Logs: int [t=1,2] [ 1 / (8-3t) ] dt

[maeveoneill]

Can someone please look over the following question and tell me where I went wrong..

For my answer I got -1/3 ln(2/5) yet the back of my text book says 1/3 ln(5/2).

evaluate the integral from 1-2 of 1/(8-3t) dt

u= 8-3t, du= -3 dt

= -1/3 |2 du/u
|1

=-1/3 [ln u] 2
1

= -1/3 (ln2- ln5)
= -1/3 ln(2/5)

 

[galactus]

That's the same thing. You're correct as is the book. The answers are equivalent.