partial fraction?

Question

solve"x/((x-1)(x^2+4)" by using partial fraction rule.give answer with explaination.

Answer 1

x / [ (x - 1)(x^2 + 4) ] = A / (x - 1) + Bx + C / (x^2 + 4)

Find a common denominator:
A / (x - 1) + Bx + C / (x^2 + 4)
= A*(x^2 + 4) / (x^2 + 4)(x - 1) + (Bx + C)(x - 1) / (x^2 + 4)(x - 1)
= (Ax^2 + 4A + Bx^2 - Bx + Cx - C) / (x^2 + 4)(x - 1)

set the numerator of the above equal to the numerator of the original expression:

(Ax^2 + 4A + Bx^2 - Bx + Cx - C) = x
(Ax^2 + Bx^2) + (-Bx + Cx) + (4A - C) = 0x^2 + 1x + 0
(A + B)x^2 + (-B + C)x + (4A - C) = 0x^2 + 1x + 0

A + B = 0
-B + C = 1
4A - C = 0

4A - B = 1
A + B = 0
5A = 1

A = 1/5
B = -1/5
C = 4/5

x / [ (x - 1)(x^2 + 4) ] =
1 / 5 (x - 1) + (-x + 4) / 5(x^2 + 4)

Answer 2

That's what I think it is:

(x) / (x-1)(x^2+4) = (A / (x-1)) + ((Bx+C) / (x^2+4)

x = A(x^2 + 4) + (Bx+C) (x-1)

substitute when x = 1:
A(1+4) = 1
thus, A = (1/5)

substitute when x = 0:
(1/5)(4) - C = 0
thus, C = (4/5)

substitute when x = 2
(1/5)(8) + (2B + (4/5)) = 2
(2B + (4/5)) = (2/5)
2B = -(2/5)
B = -(1/5)

that's it, then put the values of A, B and C in the partial fraction.

Hope I helped you.

Answer 3

first you would split it as follows:
x = A/ (x-1) + Bx+C/(x^2-4)
and then substitute x=1, x=0 and equate the coefficients of x to find A, B and C... substitute the values of A, B & C to get the answer.....
is that helpful enough??
i did partial fractions years ago but from what i remember thats how u do it....

Answer 4

2x² + 5x - 12 = (2x - 3)(x + 4), 2 non-repeated linear factors So seek for a decomp of the type 11/(2x² + 5x - 12) = A/(2x - 3) + B/(x + 4). Multiply with the aid of by utilising (2x - 3)(x + 4) to get 11 = A(x + 4) + B(2x - 3). There are a pair of methods of finding A and B. With non-repeating linear factors, we are able to easily enable x be equivalent to the roots of the quadratic. if x = -4, you get 11 = B(-8-3) = -11B ==> B = -a million if x = 3/2, you get 11 = A(3/2 + 4) = (11/2)A ==> A = 2. The decomposition is subsequently 11/(2x² + 5x - 12) = 2/(2x - 3) - a million/(x + 4). even nevertheless, you need to evaluate coefficients on the left and top. 11 = (A + 2B)x + (4A - 3B) giving A + 2B = 0, and 4A - 3B = 11. you will get a similar constants A and B.

Answer 5

x/((x - 1)(x^2 + 4)) = (A/(x - 1)) + (B/(x^2 + 4))

x = A(x^2 + 4) + B(x - 1)
2i = A((2i)^2 + 4) + B(2i - 1)
B = (4 - 2i)/5
B = (4 - x)/5
B = (-x + 4)/5
The reason we can put x in the place of 2i, is because we set x to 2i at the beginning of the work.

x = A(x^2 + 4) + B(x - 1)
1 = A(1^2 + 4) + B(1 - 1)
1 = A(1 + 4)
1 = 5A
A = (1/5)

x/((x - 1)(x^2 + 4)) = ((1/5)/(x - 1)) + ((-x + 4)/5)/(x^2 + 4))
x/((x - 1)(x^2 + 4)) = (1/(5(x - 1))) + ((-x + 4))/(5(x^2 + 4)))
x/((x - 1)(x^2 + 4)) = (1/(5(x - 1))) - ((x - 4)/(5(x^2 + 4)))

ANS : (1/(5(x - 1))) - ((x - 4)/(5(x^2 + 4)))

Answer 6

(4-x)/(5(x^2+4)) + 1/(5(x-1))

I am deliberately not giving you the working. Do it and check your answer.

Answer 7

what's it equal to?