Question

I am doing partial fraction evaluating integral(1/(x^3-x^2) and cant figure out the constant values for A B and C since 1=A(x^2)(x-1)+B(x)(x-1)+C(x-1), setting x=0 i get C=-1 but A and B i can't figure out. detailed help would be fantastic. thanks

Answer 1

\[\frac{1}{(x^3-x^2)}\]
\[\frac{1}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}\]

Answer 2

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

Answer 3

where, if any, did i mess it?

Answer 4

so don't use the faster method for this type of fraction, my prof didn't spend much time on the method you're showing
you look good so far

Answer 5

the issue is with the multiplicity of x ....

Answer 6

http://www.wolframalpha.com/input/?i=int+1%2F%28x^3-x^2%29

Answer 7

a nd b are -1 and c 1

Answer 8

in other words; when x=1; C=1 is the only option we can muster; then lets use that to zero out the others
\[1=x^3(A+C)+x^2(B-A)-Bx\]
1=x^3(A+1)+x^2(B-A)-Bx

1=x^3(-1+1)+x^2(B--1)-Bx

1=x^3(-1+1)+x^2(-1--1)--1x

Answer 9

we have to kinda combine both methods

Answer 10

stop here and solve C, for x=1
\[{1}={A(x^2)(1-1)}{}+B(1-1)(x){}+{C(x^2)(x)}{}\]
\[{1}={C}\]

then continue to the relating coeefs; with C=1

Answer 11

does that make sense?

Answer 12

ok yeah i've got what your saying. since we can determine C and it's constant we can plug it in to the expanded form

Answer 13

In your first calculation combining the partial fraction terms you've made a mistake.

Not 1 = Ax^2(x-1) + Bx(x-1) + Cx^3 ; but just
1 = Ax(x-1) + B(x-1) + Cx^2

Now you do have a constant term in A, B, C and the system can be solved.

Answer 14

james how do you get Ax not A(x^2)

Answer 15

because the denominator is not x^3(x-1) but only x^2(x-1)

Answer 16

right so when the fractions are separated you get (A/x) + (B/x^2) + C/(x-1) multiply through by the common denominator and you get A(x^2)(x-1) + B(x)(x-1) + Cx^3

Answer 17

No, NOT Ax^2(x-1). Look at the denominator. you went from A/x to something/x^2(x-1) so you need to multiple A by only x(x-1). Same goes then for the other terms

Answer 18

agreed?

Answer 19

the original denominator of the integral is x^3-x^2 which is what i'm multiplying through why would i only muliltiply through by x^2-x

Answer 20

To reiterate
1/(x^2-x) = A/x + B/x^2 + C/(x-1)
= ( Ax(x-1) + B(x-1) + Cx^2 ) / x^2(x-1)

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

and so on from here.

Answer 21

Sign error ... 1 = 1 = (A+C)x^2 + (B-A)x - B

So, explicitly B = -1, A = -1, C = 1

Thus
1/(x^3 - x) = -1/x - 1/x^2 + 1/(x-1)

This can be checked by recombining the RHS
RHS = ( -x(x-1) - (x-1) + x^2 )/(x^3-x) = 1/(x^3-x)

Answer 22

To reiterate 1/(x^2-x)<-------- not the right denominatior it's [(x^3)-(x^2] for yours you get two linear functions and would only need and A and a B, A/x and B/(x-1)

Answer 23

sorry 1/(x^3 - x^2) throughout. All of the calculation are correct.

Answer 24

I.e., 1/(x^3 - x^2) = -1/x - 1/x^2 + 1/(x-1)

Answer 25

spent enough time on this -- hope you agree. would appreciate a 'good answer'

Answer 26

ok ive got you i was multiplying through with the extra x uinder the A my bad good job