Question

integrate (x^2+8x+128)/(x^3-64)

Answer 1

can you factorise (x^2+8x+128) ?

Answer 2

I've got it mostly worked out but I keep getting the answer wrong

Answer 3

I factored the bottom and did fraction decomp

Answer 4

can you tell me what factors did you get for the numerator ?

Answer 5

\[x ^{2}+8x+128=\frac{ A }{ x }+\frac{ B }{ (x+8) }+\frac{ C }{ (x-8) }\]

Answer 6

that will be x^3 -64 x

not x^3-64

Answer 7

that's what it says, I didn't type it corrently

Answer 8

x(x+8)(x-8) = x^3-64x

but the Q has x^3-64

Answer 9

oh, cool, the Q has x^3-64x ??

Answer 10

yes, so it works out nicely

Answer 11

yeah, what values u got for A,B,C

Answer 12

A=2, B=-1 and C=0?

Answer 13

C = 0 ?
no....how ?

Answer 14

lol, that's where I figured I was messing up

Answer 15

\[A+B+C=1\]
\[-8B+8C=8\]
64A=128

Answer 16

A=2

Answer 17

\[B+C=-1\]
-B+8C=8

Answer 18

-8B

Answer 19

to get C, you need to put x=8
in
x^2+8x+128 = A (x+8)(x-8) + Bx(x-8) +C (x(x+8))
got that^ ???

Answer 20

yeah, I was solving it differently but your way is probably easier

Answer 21

then try it :)

Answer 22

especially since mine didn't work

Answer 23

C=2

Answer 24

+2 or -2 ?

Answer 25

I know it should be -2 but I'm not sure where the negative came from

Answer 26

yeah, its +2 only

Answer 27

so it is positive?

Answer 28

oh, it should be +2 only

Answer 29

what about A and B ?
to get A, you need to put x=0
in
x^2+8x+128 = A (x+8)(x-8) + Bx(x-8) +C (x(x+8))

to get B, you need to put x=-8
in
x^2+8x+128 = A (x+8)(x-8) + Bx(x-8) +C (x(x+8))

Answer 30

C= +2 is correct, if you had any doubts...

Answer 31

So A and C are 2?

Answer 32

A is not 2

Answer 33

128 = A (-64)

Answer 34

ohhhhhh

Answer 35

so, B= ?

Answer 36

So B is +1, not -1

Answer 37

Awesome, I got the answer now

Answer 38

yup
u got it all

Answer 39

Thank you!

Answer 40

welcome :)