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

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

please it is urgent(Headbang)

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- Apr 24th 2009, 06:59 AMmohamedsafypartial fraction
1/[(X^2+1)(X+2)^2]

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

please it is urgent(Headbang) - Apr 24th 2009, 07:59 AMstapel
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:

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.

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 PMmohamedsafythx
really thank u that helped me soo much