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Mathematics 26 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

y partial fractions when u can find the antiderivative directly?

OpenStudy (amistre64):

partial are cooler lol

OpenStudy (anonymous):

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

OpenStudy (anonymous):

:)

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

*A(x+4)^2

OpenStudy (amistre64):

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

OpenStudy (amistre64):

then the tops equate

OpenStudy (anonymous):

ohh okay great!

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

this looking alright so far?

OpenStudy (amistre64):

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

OpenStudy (anonymous):

im working on it :)

OpenStudy (amistre64):

forgot the B4... 16A +4B +C = 1

OpenStudy (amistre64):

i get to here and draw a blank lol

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

a = 0

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

we equate equal parts in this right? x^2(A) + x(8A + B) + 16(A) +C = 0x^2 +0x +1

OpenStudy (amistre64):

(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

OpenStudy (anonymous):

yeah

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

im game .....

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

lamee

OpenStudy (amistre64):

yeah, that would ve been awesome!! lol

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thaks for your help by the way

OpenStudy (amistre64):

and a polynomial up top to liven things up ;)

OpenStudy (amistre64):

ill be takinng calc 1 over the summer in community college

OpenStudy (amistre64):

youre welcome :)

OpenStudy (anonymous):

cool cool good for you!

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