# partial fraction

• Apr 24th 2009, 06:59 AM
mohamedsafy
partial fraction
1/[(X^2+1)(X+2)^2]

1/[X(X+1)^2]

• Apr 24th 2009, 07:59 AM
stapel
Quote:

Originally Posted by mohamedsafy

To learn, in general, how to work with partial fractions, try here. Then, since you haven't been able to start either of these, I'll give you the beginning:

Quote:

Originally Posted by mohamedsafy
1/[(X^2+1)(X+2)^2]

The denominators are (x^2 + 1), (x + 2), and (x + 2)^2. So:

. . . . .$\displaystyle \frac{1}{(x^2\, +\, 1)(x\, +\, 2)^2}\, =\, \frac{Ax\, +\, B}{x^2\, +\, 1}\, +\, \frac{C}{x\, +\, 2}\, +\, \frac{D}{(x\, +\, 2)^2}$

Multiplying through gives:

. . . . .$\displaystyle 1\, =\, (Ax\, +\, B)(x\, +\, 2)^2\, +\, (C)(x^2\, +\, 1)(x\, +\, 2)\, +\, (D)(x^2\, +\, 1)$

Letting x = -2 gives:

. . . . .$\displaystyle 1\, =\, 0\, +\, 0\, +\, D(5)$

Use this to find the value of D. Then pick three other values for x to solve for the values of A, B, and C.

Quote:

Originally Posted by mohamedsafy
1/[X(X+1)^2]

The factors in the denominators will be x, (x + 1), and (x + 1)^2. So use the standard set-up:

. . . . .$\displaystyle \frac{1}{x(x\, +\, 1)^2}\, =\, \frac{A}{x}\, +\, \frac{B}{x\, +\, 1}\, +\, \frac{C}{(x\, +\, 1)^2}$

Multiplying through gives:

. . . . .$\displaystyle 1\, =\, A(x\, +\, 1)^2\, +\, Bx(x\, +\, 1)\, +\, Cx$

Obvious values for x include x = 0 (which will give you the value of A) and x = -1 (which will give you the value of C).

If you get stuck, please reply showing how far you have gotten. Thank you! (Wink)
• Apr 24th 2009, 12:40 PM
mohamedsafy
thx
really thank u that helped me soo much