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Re: HSC 2017 MX2 Integration Marathon \text{For}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{re-write as}\int x (\frac{e^{x}}{1+e^{x}})dx\\\text{Then do IBP and a substitution in the remaining integral and bob's your uncle}

Re: HSC 2017 MX2 Integration Marathon Wait what's this reserve quotient rule thing? I've never heard of this...

Re: HSC 2017 MX2 Integration Marathon Correct! Sorry if this seems like a dumb question, how do you go from line 2-3? Cause I can only do it your way by splitting the integral to \int\frac{1}{1+lnx}dx-\int\frac{x}{x(1+lnx)^{2}}dx. Then if I do the 2nd integral by IBP the remaining integrals...

Re: HSC 2017 MX2 Integration Marathon Also can work by \times\frac{x}{x}

Re: HSC 2017 MX2 Integration Marathon I don't know what closed form is but you just need to be creative

Re: HSC 2017 MX2 Integration Marathon Correct! One of those ones where you have to complete the square then do the appropriate trig sub and fiddle with a right angled triangle...

Re: HSC 2017 MX2 Integration Marathon \text{1.}\int\sqrt{-4x^{2}+3x+6}dx\\\text{2.}\int\frac{e^{x}}{1+e^{x}}dx\\\text{3.}\int\frac{xe^{x}}{1+e^{x}}dx\\\text{4.This was actually the last q in a hsc paper somewhere and involves a weird trick}\int\frac{lnx}{(1+lnx)^{2}}dx

Anyone heading into campus tomorrow. Don't be afraid to say hi. I'll be down for the business faculty welcome etc.

You lost me... Why did you go to logs and sigma?

So what you're telling me is that \prod\limits_{n=1}^{\infty}\frac{2n^{2}-n-1}{2n^{2}+n-1}\neq(\prod\limits_{n=1}^{\infty}\frac{n-1}{n+1})(\prod\limits_{n=1}^{\infty}\frac{2n+1}{2n-1})? Is there a rule kinda like a product of matrices going on here?

Sorry to bother you, would you be able to post a proof then?

Hey, 1. Find the equation for the parabola with the parameter removed 2. Find the equation of the normal at P or Q then find the other by inspection 3. Find the equation of the line PQ 4. Sub in the given point to find an expression for pq 5. Find the point of int of the normals 6. Find...

But I don't quite understand the difference between \prod\limits_{n=1}^{\infty}\frac{2n^{2}-n-1}{2n^{2}+n-1} and \lim_{x\to\infty}S_n.

Is it always 0? I don't know how to explain it for all cases but I just tried \prod\limits_{n=1}^{\infty}1-\frac{1}{n^{2}} because I think it satisfies the above condition and I got \infty\times0, so I thought it'd be 0?

So k is just any random number that satisfies 2\leq{k}<\infty? And if so you need to do this because it doesn't make sense for k to be infinity...? Don't quite understand what you mean

Here's my thinking on it. I got the same answer as Kingom but I haven't done many infinite product questions...

Btw hectic marks :)

Yeah I always forget about your reason for demand -_-

With the people and place units, like are they mini-internships i.e. do you have to find somewhere to take you on or can you still do it at mq?

Hey, With these questions you just need to think x->y->z etc. The RBA can influence the general level of interest rates within the domestic economy through altering the cash rate and also trading foreign currency or otherwise know as 'dirtying the float'. So since the price of AUD has fallen...