Posted by mousey on Saturday, October 24, 2009 at 5:12pm.
using method of partial fractions to decompose this 1/(x3)(x+3)
is 1/6(x3)1/6(x+3) correct or have i gone wrong somewhere

math  jim, Saturday, October 24, 2009 at 7:08pm
I get
1/(x3)(x+3) = A/(x3) + B/(x+3)
1 = (x+3)A + (x3)B
1 = x(A+B) + 3A 3B
> A + B = 0 > A = B
> 3A  3B = 1 > A = 1/6; B = 1/6
= (1/6)(x3)  (1/6)/(x+3)

math  mousey, Sunday, October 25, 2009 at 5:45am
thanks so it is the same thank god for that
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