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y partial fractions when u can find the antiderivative directly?
partial are cooler lol
lol i said the same thing but that is the unit we are on!
A + B + C ----- ------- -------- (x+4) (x+4)^2 (x+4)^3
correct now solve it and u get A= B= 0 :P
A(x+4)^2 ------------ (x+4)(x+4)^2
A(x+4)^2 +B(x+4)+ C 1 ---------------------- = ------- (x+4)^3 (x+4)^3 right?
i thought you had to do it where you cancel denominators and set 1= A(x+4)2 + B(x+4) +C and then solve
you tend to split them to see what you had to multiply by to get common denominators
then the tops equate
ohh okay great!
A(x+4)^2 +B(x+4)+ C = 1 A(x^2 +8x +16) +Bx +B4 +C = 1
Ax^2 +A8x +A16 +Bx +B4 +C = 1
this looking alright so far?
x^2(A) + x(8A + B) + 16(A) +C = 0x^2 +0x +1
im working on it :)
forgot the B4... 16A +4B +C = 1
i get to here and draw a blank lol
do we find a value for a variable and stick it in?
yes we need to be able to know what A B and C are equal to
a = 0
x^2A = x^2(0) A = 0 right?
16(0) +4(0) +C = 1 C = 1
we equate equal parts in this right? x^2(A) + x(8A + B) + 16(A) +C = 0x^2 +0x +1
(0)x^2 = Ax^2; A = 0 (0)x = (8A+B)x = Bx; B=0 A16 +4B + C = 1 0+0+C = 1 c = 1
so the derivative is simply the derivative of the original equation :)
ugh oh. let me get back to you i think we may have gone the wrong rout earlier would you like to work through an easier problem so i can show you the jist of it and then maybe we can come back to solving this one? feel free to say no
nvm u are right and the antiderivative is the same as the orginal solve the shorter way
im game .....
if they had given a factored form of an equation inthe bottom with different fillers, it would have been way more exciting lol
yeah like 1/(x-1)(x+2)^2
yeah, that would ve been awesome!! lol
so i see you enjoy math lol? im a soph taking calc II
thaks for your help by the way
and a polynomial up top to liven things up ;)
ill be takinng calc 1 over the summer in community college
youre welcome :)
cool cool good for you!