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It won't matter much at all.

You can drop it anytime on MyAdmin before the census date.

Thanks examine!

The faculty of science at USYD does have internal transfers within the faculty.

Last few spots remaining!

Lol. Dole is dole.

My ya basically disappeared after 2014 happened so thankfully I get start up (>1 dollar a fort night ) qualifies

Note that this was used in the context of string theory so a lot of mathematics is just taken as things appear to be : (Oh..it's -1/12 and if we assume the cessaro mean it works!! OMGZZ STRING FEORY!!) In a mathematics sense, obviously it's not rigorous at all and is wrong.

Re: MX2 Integration Marathon Make them up, famous integration problems

Re: MX2 Integration Marathon \sqrt{\dfrac{1+x}{3+x}} = \sqrt{\dfrac{1+x}{3+x}} \cdot \dfrac{\sqrt{1+x}}{\sqrt{1+x}} = \dfrac{1+x}{\sqrt{(1+x)(3+x)}} = \dfrac{1+x}{\sqrt{x^2+4x+3}} and you should be able to go from there.

Re: MX2 Integration Marathon Yes I had this going through my head and doubled checked it. Sometimes Yr8 permeates the mind.

2^{\ln x} = e^{\ln x \cdot \ln 2} = (e^{\ln x})^{\ln 2} = x^{\ln 2} Then treat ln2 as n, so what's the integral of x^n?

Re: MX2 Integration Marathon For those 99% of the time it's a rationalise the numerator problem.

Yes...but since last year. So I think it'll be a loan for you (legislation still hasn't been passed so idk).

u = \ln x \Rightarrow e^u = x \\ du = \dfrac{dx}{x} \\ I = \int 2^u e^u \mathop{du} \\ = \int (2e)^u \mathop{du} \\ = \int e^{\ln (( 2e)^u)} \mathop{du} \\ \int e^{u\ln(2e)} \mathop{du} \\ And go from there.

Use a substitution.

And yet you somehow still are into cags...

Re: MX2 Integration Marathon I remember this from 2012 hahha. Leme see if I remember how to do it... (ceebs actually)

I'm guessing you're gonna get so many cags?