Question

integral dx/x^3-1?
i already got A which is 1/3 but how can i find b and c?

Answer 1

\[\int\limits \frac{ dx }{ x ^{3}-1 } = \int\limits \frac{ A }{ x-1}+\frac{ B }{x ^{2}-1 }+\frac{ C }{ x ^{3}-1 }\]
is this your equation?

Answer 2

its \[\int\limits \frac{ dx }{ x^3-1 }\]
and the answer from my calculus book shows me with a inverse tangent

Answer 3

did you setup up A+B+C the same way that I did it?

Answer 4

FYI, x^3-1= ( x-1)*(x^2+x+1)

Answer 5

the factorization you have is incorrect, and hence you will not get correct values of B AND C

Answer 6

you cant factor x^2+x+1

Answer 7

it does not have real roots

Answer 8

but u have to complete the square right?

Answer 9

\[\int\limits\ \frac{ dx }{ (x-1)(x ^{2} +x+1)} = \int\limits\ \frac{ A }{ (x-1) }+\frac{ Bx+C }{ (x ^{2}+x+1) }dx\]
how does this look?

Answer 10

thats correct

Answer 11

try it out, should work

Answer 12

ok let's see

Answer 13

if not try BX^2+Cx+D

Answer 14

i just did that but i am trying to find B and C. i got A as 1/3

Answer 15

first try with @VeritasVosLiberabit , suggestion, i think you should be able to find b and c, if not try what i suggested with B,C and D

Answer 16

\[1=(x ^{2}+x+1)A + (x-1)(Bx+C)\]
for x = 1
\[1=3A, A=\frac{ 1 }{ 3 }\]
\[1=Ax ^{2}+Ax+A+Bx ^{2}+Cx-Bx-C\]
\[1=(A+B)x ^{2}+(A-B+C)x+A+C\]

Answer 17

put x=0 and see what you get

Answer 18

yeah I think we can solve for C with x=0

Answer 19

sorry, you need to try , i will verify your work, but not give you exact answer

Answer 20

\[1=\frac{ 1 }{ 3 }+C, C=\frac{ 2 }{ 3 }\]

Answer 21

great now take any value of x, say x=-1 and sub values of A and C to get B

Answer 22

so with those two values you should be able to get B

Answer 23

just to let you know A,B and C can be obtained using various methods

Answer 24

I need to brush up on my factoring

Answer 25

c=-2/3 and b=-1/3. that is wut i got

Answer 26

does that help

Answer 27

yeah it does XD. thank u

Answer 28

yw