Question

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

Answer 1

y partial fractions when u can find the antiderivative directly?

Answer 2

partial are cooler lol

Answer 3

lol i said the same thing but that is the unit we are on!

Answer 4

:)

Answer 5

A + B + C
----- ------- --------
(x+4) (x+4)^2 (x+4)^3

Answer 6

correct now solve it and u get A= B= 0 :P

Answer 7

A(x+4)^2
------------
(x+4)(x+4)^2

Answer 8

A(x+4)^2 +B(x+4)+ C 1
---------------------- = -------
(x+4)^3 (x+4)^3


right?

Answer 9

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

Answer 10

*A(x+4)^2

Answer 11

you tend to split them to see what you had to multiply by to get common denominators

Answer 12

then the tops equate

Answer 13

ohh okay great!

Answer 14

A(x+4)^2 +B(x+4)+ C = 1
A(x^2 +8x +16) +Bx +B4 +C = 1

Answer 15

Ax^2 +A8x +A16 +Bx +B4 +C = 1

Answer 16

this looking alright so far?

Answer 17

x^2(A) + x(8A + B) + 16(A) +C = 0x^2 +0x +1

Answer 18

im working on it :)

Answer 19

forgot the B4...

16A +4B +C = 1

Answer 20

i get to here and draw a blank lol

Answer 21

do we find a value for a variable and stick it in?

Answer 22

yes we need to be able to know what A B and C are equal to

Answer 23

a = 0

Answer 24

x^2A = x^2(0) A = 0 right?

Answer 25

16(0) +4(0) +C = 1

C = 1

Answer 26

we equate equal parts in this right?

x^2(A) + x(8A + B) + 16(A) +C = 0x^2 +0x +1

Answer 27

(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

Answer 28

yeah

Answer 29

so the derivative is simply the derivative of the original equation :)

Answer 30

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

Answer 31

nvm u are right and the antiderivative is the same as the orginal solve the shorter way

Answer 32

im game .....

Answer 33

if they had given a factored form of an equation inthe bottom with different fillers, it would have been way more exciting lol

Answer 34

yeah like 1/(x-1)(x+2)^2

Answer 35

lamee

Answer 36

yeah, that would ve been awesome!! lol

Answer 37

so i see you enjoy math lol? im a soph taking calc II

Answer 38

thaks for your help by the way

Answer 39

and a polynomial up top to liven things up ;)

Answer 40

ill be takinng calc 1 over the summer in community college

Answer 41

youre welcome :)

Answer 42

cool cool good for you!