integration question (1 Viewer)

polythenepam

Member
Joined
Jun 10, 2005
Messages
57
Gender
Female
HSC
2005
hey i need help
the question is
integrate :
(Ln(x^2-4))/(x^2-4)^(1/2) dx

before that it asks to show the integral of
x/(x^2-4)dx = (x^2-4)^1/2
 

polythenepam

Member
Joined
Jun 10, 2005
Messages
57
Gender
Female
HSC
2005
yeeh im sure.. to clarify :
the integral of :
"[Ln of (x squared minus 4)] divided by [square root of (x squared minus 4)]"
 

HayleeKate

Member
Joined
Mar 3, 2005
Messages
237
Location
Annandale
Gender
Female
HSC
2005
I beg to differ? I've checked and re-checked several times because I'd back yourself over me, Slide, BUT...
u=sqrt(x^2-4)
du=1/2.2x.(x^2-4)^(1/2)
=x/sqrt(x^2-4)
doesnt it? where does your 2x disappear too? Do I not understand?

Also, considering the first part of the question asks you to show the integral of x/sqrt(x^2-4)... i'd suggest you ought to be using this?
 
Last edited:

haboozin

Do you uhh.. Yahoo?
Joined
Aug 3, 2004
Messages
708
Gender
Male
HSC
2005
mmmm currently ngai is top of the state for last yr....
so i would trust everything he says :p

also your integration cant be solved with high school math...


Integrate



Result
 
Last edited:

martin

Mathemagician
Joined
Oct 15, 2002
Messages
75
Location
Brisbane
Gender
Male
HSC
2002
for the previous integral I don't think the answer is right

int(x/(x^2-4))dx = ?

let u=x^2-4, then du = 2 x dx so x dx= du/2

so int(x/(x^2-4))dx
= int(1/2 1/u)du
= 1/2 ln(u) + C
= 1/2 ln(x^2-4) + C
= ln(sqrt(x^2-4)) + C

But how this helps us for the desired integral I don't know. Are you sure you typed it in right or were there any limits provided?
 
Last edited:

polythenepam

Member
Joined
Jun 10, 2005
Messages
57
Gender
Female
HSC
2005
hmm yeeh there were limits
8^1/2 and 5^1/2
i dont think that makes a difference tho
well yeeh my friend asked me this question so maybe he took it down wrong.. hopefully it's wrong.. no wonder its so screwed
anyway thnx for all ur help
 

FinalFantasy

Active Member
Joined
Jun 25, 2004
Messages
1,179
Gender
Male
HSC
2005
polythenepam said:
hmm yeeh there were limits
8^1/2 and 5^1/2
i dont think that makes a difference tho
well yeeh my friend asked me this question so maybe he took it down wrong.. hopefully it's wrong.. no wonder its so screwed
anyway thnx for all ur help
limits make a huge difference... some ppl just post integration questions without putting the full question in, and we waste lots of time doing it and gain no result lol:\
 

Antwan23q

God
Joined
Sep 12, 2004
Messages
294
Location
bally
Gender
Undisclosed
HSC
2006
polythenepam said:
before that it asks to show the integral of
x/(x^2-4)dx = (x^2-4)^1/2
Geez, that is totaly wrong!
integration of x/(x^2-4) is.... 1/2ln|x^2-4|

the derative of (x^2-4)^1/2 = x/((x^2-4)^1/2))


u obviously forgot the the square root infront of the (x^2-4)

that should help i guess
 

polythenepam

Member
Joined
Jun 10, 2005
Messages
57
Gender
Female
HSC
2005
yeeh they make a difference to the overall answer but u cant use those limits unless u can get the indefinite integral first
ahh well looks like u cant integrate it anyway
 

~ ReNcH ~

!<-- ?(°«°)? -->!
Joined
Sep 12, 2004
Messages
2,493
Location
/**North Shore**\
Gender
Male
HSC
2005
haboozin said:
mmmm currently ngai is top of the state for last yr....
so i would trust everything he says :p

also your integration cant be solved with high school math...


Integrate



Result
Where did you get that result from?...do you have an integral calculator or something?
 

haboozin

Do you uhh.. Yahoo?
Joined
Aug 3, 2004
Messages
708
Gender
Male
HSC
2005
~ ReNcH ~ said:
Where did you get that result from?...do you have an integral calculator or something?


yea math.com
this integral is beyond our syllabus
 

HayleeKate

Member
Joined
Mar 3, 2005
Messages
237
Location
Annandale
Gender
Female
HSC
2005
this question is a filthy wench.
I tried int by parts first off, thought that would work after the first part of the question, but ended up going around in circles [but not enough to get a recurrence].
The substitution never works out properly. I cant find a similar example in any of my textbooks either.

PS Slide, I was scared to doubt you. It was a much nicer question your method anyway!
 

ngai

Member
Joined
Mar 24, 2004
Messages
223
Gender
Male
HSC
2004
FinalFantasy said:
limits make a huge difference... some ppl just post integration questions without putting the full question in, and we waste lots of time doing it and gain no result lol:\
I AGREE...which is why i feed questions into maple before trying myself

polythenepam said:
hmm yeeh there were limits
8^1/2 and 5^1/2
i dont think that makes a difference tho
well yeeh my friend asked me this question so maybe he took it down wrong.. hopefully it's wrong.. no wonder its so screwed
anyway thnx for all ur help
maple still cant do it, so im pretty sure u wont need to do it in the hsc
 

who_loves_maths

I wanna be a nebula too!!
Joined
Jun 8, 2004
Messages
600
Location
somewhere amidst the nebulaic cloud of your heart
Gender
Male
HSC
2005
Originally Posted by polythenepam
hey i need help
the question is
integrate :
(Ln(x^2-4))/(x^2-4)^(1/2) dx

before that it asks to show the integral of
x/(x^2-4)dx = (x^2-4)^1/2
i just had this question asked of me last week by a fellow student who was looking at past papers from a Catholic school's trial. (i think he got it through a tutor or something).
anyways, the point this question is misplaced at the HSC level and the integral cannot be found using high school mathematics, even if the integral is a definite integral. unless you know what Polylogarithmic functions are, then it can't be done.

if i remember correctly you are suppose to find its definite integral between sqrt(5) and sqrt(8)... i think... and you're suppose to express it in exact form ---> which leads me to believe the integral is actually that of x(ln(x^2-4))/sqrt(x^2-4) , they just forgot the extra 'x'.


P.S. i don't think that the definite integral for ln(x^2-4)/sqrt(x^2-4) between sqrt(5) and sqrt(8) can even be expressed in exact form. (certainly not interms of root extractions or logarithms and exponential values).
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top