Question

Is there a method to use besides trig sub for this problem? If not, How can I solve it with T.S.?

Answer 1

\[\int\limits_{}^{}1/(3x ^{2}-81)\]

Answer 2

it doesn't factor

Answer 3

really?

Answer 4

partial fraction won't work

Answer 5

3(x^2-27)

Answer 6

but that does not help

Answer 7

Yea, I did that to get \[1/3\int\limits_{}^{}1/(x ^{2}-27)\]

Answer 8

The way I do trig sub is to make a triangle

Answer 9

Some trig sub problem can be done easily instead with polar coordinate but not this one

Answer 10

And in this scenario the hyp is x, adjacent is \[\sqrt(27)\] and the opposite is \[\sqrt(x^2-27)\] ?

Answer 11

Not 100% positive about those sides

Answer 12

why not partial fractions?
\[\frac{1}{3}\int \frac{dx}{(x+3\sqrt{3})(x-3\sqrt{3})}\]

Answer 13

Yea that would work but I wouldn't have seen it on a test lol

Answer 14

can't imagine another way though.

Answer 15

Thanks, how about trig sub approach?

Answer 16

maybe hyperbolic? look in a book for that

Answer 17

http://www.wolframalpha.com/input/?i=int%3A+1%2F%283x^2-81%29

Answer 18

I think we might sub Sec(t) for x

Answer 19

Yea I think thats what wolfram did there but I haven't even learned hyperbolic yet

Answer 20

yeah fine but look at the answer. you get log something plus log something so i am thinking partial fractions is the way to go

Answer 21

yay, partial

Answer 22

Okie dokes

Answer 23

Thanks!

Answer 24

not that big of a deal i think. write
\[(x+3\sqrt{3})A+(x-3\sqrt{3})B=1\]

Answer 25

what about the 1/3?

Answer 26

Does it just hang out till we get the answer?

Answer 27

the integration*

Answer 28

put
\[x=3\sqrt{3}\] get
\[ 6\sqrt{3}A=1\]
\[A=\frac{1}{6\sqrt{3}}\] etc.

Answer 29

yeah just leave the 1/3 outside

Answer 30

and don't get married to wolfram's answer since there are many ways to write a log

Answer 31

ok lemme take a swing at this with partial, thanks

Answer 32

i will be without looking that if
\[A=\frac{1}{6\sqrt{3}}\] then
\[B=-\frac{1}{6\sqrt{3}}\] t

Answer 33

enjoy