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split into partial fractions and integrate. (x-3)-4/(x-2)

First Class Honours (Pure Mathematics) at USYD, prepare to be destroyed with mathematical knowledge..

about 3500 if numbers are consistent with last year

so you give an example of when u use delta instead?

ahhh damn seanieg coming in with the hammer :)

just shows my eloquence and flair ;)

function of a function. let u=x/2

the <a href="http://www.codecogs.com/eqnedit.php?latex=\delta" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\delta" title="\delta" /></a> is just delta which means "change in" in greek. I thought it was the "correct" letter to be used and people just used d for ease? or am i...

love it :) get repped for fucking up ahah

i guess i had kind of used a fact to prove a fact. should have done it the other way like nightweaver

i don't find it complicated at all haha

hahahahah lol i fail at failing

u need to prove that homeboy

tan(\alpha )=y\\ \alpha =tan-1(y)\\ \delta \alpha /\delta y=1/1+y^2\\ \delta y/\delta \alpha =1+y^2\\ =1+tan^2(\alpha )\\ =sec^2\alpha dayum :(

same shit different smell, test the derivative of your answer to see if its correct

u will have a simple substitution, du=cosxdx. which will leave the integral of u^3-u^5. for your other integral, make sure you evaluate the unintegrated term from 0 to ln2 as well

change one of the cos^2 x to 1-sin^2x, and let u=sinx

for efficiency and simplicity, i would use ln(x) instead of loge(x), but each to their own :)

are u doing accelerated maths?

<a href="http://www.codecogs.com/eqnedit.php?latex=\inline s=\frac{d}{t}\\ t=\frac{d}{s}\\ \\ downstream\\ \\ 3=\frac{d}{a@plus;b}~where~a~is~rowspeed~and~b~is~riverspeed\\ \\ upstream\\ 4=\frac{d}{a-b}\\ \\ simultaneously\\ \frac{3}{4}=\frac{d(a-b)}{d(a@plus;b)}\\...