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dauswq
x^2-4 (x-2)(x+2)
using partial fraction:
x+1 = A + B (2) (x-2)(x+2) (x-2) (x+2) x+1=A(x+2) + B(x-2) when x=-2 -1=-4B B = 1/4 when x=2 3=4A A=3/4 substitute back in (2) x+1 = 3 + 1 (x-2)(x+2) 4(x-2) 4(x+2) integration of (1)= x^3 + 3 ln (x-2) + ln (x+2) + C 3 4 4
2. (1) x^2(x^2-4) + x + 1 = x^2 + x+1 x^2-4 (x-2)(x+2) u ... lavender_aqua Post at 22-3-2012 00:11
terime kasih n tahniah sebab jawab jugak.. aritu tak perasan ke ade kuiz aqu ... midory Post at 22-3-2012 00:31
sama2... salah masuk thread kuiz... lavender_aqua Post at 22-3-2012 00:49
thread yg dh merge tu eh... midory Post at 22-3-2012 00:54
aah.. sbb nak tau jwpn last week... lavender_aqua Post at 22-3-2012 00:56
~Tahap Soalan 8~ Sedang
PENYELESAIAN
1. 2.
1. let u=ln(ln x) y=ln u dy/du=1/u since u=ln(ln x) let v=ln x --> dv/dx=1/x so that u=ln ... lavender_aqua Post at 21-3-2012 23:58
Ehhh takde orang nak jawab ke? Saya ni kalau tengok soalan calculus, geram... rasa nak solve cepat2. ... mnm77 Post at 20-3-2012 22:08
SOALAN 9
Hint: Tanya fasilitator (dauswq, midory) dengan pm kami serentak...Satu hint akan mengurangkan markah sebanyak -5.
Initial conditions (IC) diberi tak? Tak diberi kan? ....mana tau kot-kot la ada diberi tapi s ... mnm77 Post at 26-3-2012 11:52
takde mnm... bg penyelesaian am je.. midory Post at 26-3-2012 15:08
alamak, ter'advanced'... biarlah... lavender_aqua Post at 26-3-2012 16:03
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