Question

Evaluate integral 3x lnx/ sqrt (x^2-64) dx

Answer 1

the answer key says the last step is tan ^-1 not sec, i'm not sure, can someone check this over?

Answer 2

Earlier in the problem, when you were plugging everything in, I think the integral becomes \[\huge 24\int\limits \sec^2 \theta \ln ( \sec \theta ) d \theta\]
When you plugged in the 8tan(theta) it looks like u accidentally plugged in tan(theta) instead. You can correct me if I'm wrong though, lot of work to this problem :3 lol

As far as the arctan thing goes, Hmmmmmmmmmmmmmm....

Answer 3

no, that 8 i moved it outside
ok..my numbers are kinda messed up lol
the first line suppose to be a 3
and 2nd line is 24

Answer 4

yea, this assignment is tough, but these ques are still consider to be fairly ok, there are a few of them that took me more than an hour to finish..

Answer 5

\[\large x=8\sec \theta \qquad dx=8\sec \theta \tan \theta d \theta\]
\[\huge \int\limits \frac{3x \ln x}{\sqrt{x^2-64}}dx\]
\[\large \rightarrow \quad \int\limits \frac{3(8\sec \theta) \ln (8\sec \theta)}{8\tan \theta}(8\sec \theta \tan \theta d \theta)\]
Did I fill those in correctly? :o

Answer 6

oh..i get what u are saying

Answer 7

i missed an 8 in the denominator.
then the constant should just be 24 instead of 192

Answer 8

and how about my sec ^-1 x at the end? is that correct?

Answer 9

It looks correct :\ lemme take another look hmm

Answer 10

ops i forgot to choose the best response for the last post xD

Answer 11

ok thanks!

Answer 12

Yah it seems like the last term should be an arcsecant, hmm I'm not sure what they're doing there. :p

Answer 13

ok thanks zepdrix :D