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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

:)
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
*A(x+4)^2
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
yeah
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
lamee
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!

Not the answer you are looking for?

Search for more explanations.

Ask your own question