Question 1

[syn14]

Wasn't it -1/3 (2* log e l 3 - x l + log e l 3+x l ) ??

This. The partial fraction A/(3 ‐x) + B(3+x) got me A = 2/3 and B =  ‐1/3. Integrate from there for the above.

[jane1234]

Wasn't it -1/3 (2* log e l 3 - x l + log e l 3+x l ) ??

This. The partial fraction A/(3 ‐x) + B(3+x) got me A = 2/3 and B =  ‐1/3. Integrate from there for the above.

*massive sigh of relief*

I really should stop looking at what other people got...

[golden]

Wasn't it -1/3 (2* log e l 3 - x l + log e l 3+x l ) ??

This. The partial fraction A/(3 ‐x) + B(3+x) got me A = 2/3 and B =  ‐1/3. Integrate from there for the above.

*massive sigh of relief*

I really should stop looking at what other people got...

Agreed. I was VERY relieved.

It's ok if we leave it in two separate parts right?

[mattshen]

i'm not too sure about this. because i got something like 1/6 loge abs(x+3/x-3) - 1/2 loge abs (9-x^2) :S 

[jane1234]

Wasn't it -1/3 (2* log e l 3 - x l + log e l 3+x l ) ??

This. The partial fraction A/(3 ‐x) + B(3+x) got me A = 2/3 and B =  ‐1/3. Integrate from there for the above.

*massive sigh of relief*

I really should stop looking at what other people got...

Agreed. I was VERY relieved.

It's ok if we leave it in two separate parts right?

No idea, but they didn't ask for it simplified or in a certain form.

I thought it would be safer leaving it as it was - less chance of silly mistakes...

And @mattshen I think it depended on how you split the fraction - it might be equivalent though...

[lateralus]

i'm not too sure about this. because i got something like 1/6 loge abs(x+3/x-3) - 1/2 loge abs (9-x^2) :S

Same. Heaps of other people did too, and someone said it will get you 3/3 so don't worry

[mattshen]

i'm not too sure about this. because i got something like 1/6 loge abs(x+3/x-3) - 1/2 loge abs (9-x^2) :S

Same. Heaps of other people did too, and someone said it will get you 3/3 so don't worry

phew thank you mate

[BoredSatan]

yeah there are 2 methods

method 1:
Split the integral into x/(9-x^2) +x/(9-x^2)
and then partial fraction the second part
i did this and got the one with 1/6log....

method 2:
use partical fractions from the start
ie (x+1)/(3-x)(3+x)

you get the same answer as itute.

Both should be accepted

[funkyducky]

Both answers are legit, it depends on your approach - either doing partial fractions on the original expression or splitting it and doing partial fractions on the bit with 1 as the numerator.

[4231]

i got something like -1/2lnl9-x^2l+1/6lnl3-xl-1/6lnl3+xl

Yeh i got the same thing, and am pretty sure its right. if u differentiate it on ur calc it gives you the answer. An im guessing to get that answer u split it up and did a u sub and partial fractions? its not the simplest answer but im pretty confident its right

[BoredSatan]

i got something like -1/2lnl9-x^2l+1/6lnl3-xl-1/6lnl3+xl

Yeh i got the same thing, and am pretty sure its right. if u differentiate it on ur calc it gives you the answer. An im guessing to get that answer u split it up and did a u sub and partial fractions? its not the simplest answer but im pretty confident its right

yeah i split then used partial fractions.

im just a little worried i didnt combine the logs. but hopefully it should be fine

[AleksIlia]

i got something like -1/2lnl9-x^2l+1/6lnl3-xl-1/6lnl3+xl

Yeh i got the same thing, and am pretty sure its right. if u differentiate it on ur calc it gives you the answer. An im guessing to get that answer u split it up and did a u sub and partial fractions? its not the simplest answer but im pretty confident its right

yeah i split then used partial fractions.

im just a little worried i didnt combine the logs. but hopefully it should be fine

Hmm, I gave my answer as the sum of two logs, do you guys reckon we will still be alright, seeing as they didn't ask for a specific form and the assessment reports usually say '' equivalent forms were accepted.''?

[BoredSatan]

i got something like -1/2lnl9-x^2l+1/6lnl3-xl-1/6lnl3+xl

Yeh i got the same thing, and am pretty sure its right. if u differentiate it on ur calc it gives you the answer. An im guessing to get that answer u split it up and did a u sub and partial fractions? its not the simplest answer but im pretty confident its right

yeah i split then used partial fractions.

im just a little worried i didnt combine the logs. but hopefully it should be fine

Hmm, I gave my answer as the sum of two logs, do you guys reckon we will still be alright, seeing as they didn't ask for a specific form and the assessment reports usually say '' equivalent forms were accepted.''?

yeah should be fine..

[BubbleWrapMan]

VCAA tends to comment, at least on Methods reports, that students tend to proceed beyond what is explicitly asked for in a question and incorrectly simplify as a result. So I'm thinking the form doesn't matter as long as it's an antiderivative.

[Andiio]

Wasn't it -1/3 (2* log e l 3 - x l + log e l 3+x l ) ??

This. The partial fraction A/(3 ‐x) + B(3+x) got me A = 2/3 and B =  ‐1/3. Integrate from there for the above.

*massive sigh of relief*

I really should stop looking at what other people got...

Agreed. I was VERY relieved.

It's ok if we leave it in two separate parts right?

No idea, but they didn't ask for it simplified or in a certain form.

I thought it would be safer leaving it as it was - less chance of silly mistakes...

And @mattshen I think it depended on how you split the fraction - it might be equivalent though...

Scared the hell out of me, my answer was the one with 1/6