Find the partial fraction of the following:

Question

I have worked this out myself but it's not the same as my friend is getting so we need someone else to give it ago please, again please show your steps:

 

Find the partial fraction of the following:

x/(x+1)(x²+2x+6)

 

Thank you

Answer 1

x/[(x + 1)(x² + 2x + 6)] = A/(x + 1) + (Bx + C)/(x² + 2x + 6)

 

Multiply the equation by (x + 1)(x² + 2x + 6) to get:

 

x = A(x² + 2x + 6) + (Bx + C)(x + 1)

 

x = A(x² + 2x + 6) + Bx² + Bx + Cx + C

 

x = (A + B)x² + (2A + B + C)x + (6A + C)

 

0x² + 1x + 0 = (A + B)x² + (2A + B + C)x + (6A + C)

 

Equating coefficients of like powers, we have:

 

A + B = 0

2A + B + C = 1

6A + C = 0

 

Since A + B = 0, A = -B

 

Replace A by -B in the other two equations to obtain:  -B + C = 1

                                                                                              -6B + C = 0

 

Subtracting these equations, we get 5B = 1.  So, B = 1/5

 

Therefore, A = -B = -1/5 and C = -6A = 6/5

 

The partial fraction decomposition is:

 

(-1/5)/(x + 1) + (1/5)x/(x² + 2x + 6) + (6/5)/(x² + 2x + 6)

Answer 2

To find the partial fractions, write this as

 

  A            Bx + C

------- + --------------

x + 1     x² +2x + 6

 

 

Then do the sum of the two fractions to get the following:

 

Ax² + 2Ax + 6A + Bx² + Cx + Bx + C

---------------------------------------------

  (x+1) (x² +2x + 6)

 

 

Collect the terms:

 

(A + B)x² + (2A + B + C)x + (6A + C)

---------------------------------------------

  (x+1) (x² +2x + 6)

 

 

Since the numerator has to be x, this means that:

 

A + B          = 0

2A + B + C  = 1

6A + C        = 0

 

 

Solve this system and you get:

 

A = -1/5

B = 1/5

C = 6/5

 

So the partial fraction decomposition is:

 

 -1/5        (1/5)x + 6/5
------- + ------------------    =
x + 1      x² +2x + 6

 

 

  1    (      -1              x + 6          )

----   (  ------- + ------------------    )

  5    (    x + 1      x² +2x + 6      )