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anonymous
 5 years ago
i need to find the antiderivative of 1/(x+4)^3 using the method of partial fractions
anonymous
 5 years ago
i need to find the antiderivative of 1/(x+4)^3 using the method of partial fractions

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y partial fractions when u can find the antiderivative directly?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0partial are cooler lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol i said the same thing but that is the unit we are on!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0A + B + C    (x+4) (x+4)^2 (x+4)^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0correct now solve it and u get A= B= 0 :P

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0A(x+4)^2  (x+4)(x+4)^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0A(x+4)^2 +B(x+4)+ C 1  =  (x+4)^3 (x+4)^3 right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i thought you had to do it where you cancel denominators and set 1= A(x+4)2 + B(x+4) +C and then solve

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you tend to split them to see what you had to multiply by to get common denominators

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0A(x+4)^2 +B(x+4)+ C = 1 A(x^2 +8x +16) +Bx +B4 +C = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Ax^2 +A8x +A16 +Bx +B4 +C = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this looking alright so far?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x^2(A) + x(8A + B) + 16(A) +C = 0x^2 +0x +1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0forgot the B4... 16A +4B +C = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i get to here and draw a blank lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do we find a value for a variable and stick it in?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes we need to be able to know what A B and C are equal to

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x^2A = x^2(0) A = 0 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.016(0) +4(0) +C = 1 C = 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we equate equal parts in this right? x^2(A) + x(8A + B) + 16(A) +C = 0x^2 +0x +1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so the derivative is simply the derivative of the original equation :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ugh 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nvm u are right and the antiderivative is the same as the orginal solve the shorter way

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if they had given a factored form of an equation inthe bottom with different fillers, it would have been way more exciting lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah like 1/(x1)(x+2)^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, that would ve been awesome!! lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i see you enjoy math lol? im a soph taking calc II

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thaks for your help by the way

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and a polynomial up top to liven things up ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ill be takinng calc 1 over the summer in community college

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cool cool good for you!
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