1 maths Q (What's wrong?)

Interesting, the point is the LCM of the denominator of the RHS of your 'Let' equation is (x²-1), not (x-1)(x+1)²,
which is differ from LHS.
So, you are using a wrong method.

(Instead, try the following)
Let x/[(x+1)(x²-1)]=(Bx+C)/(x²-1)+(Dx+E)/(x+1)²__________________x=(Bx+C)(x+1)+(Dx+E)(x-1)__________________x=(B+D)x²+(-B+C+D+E)x+(-Cx+E)By comparing the coefficientsB+D=0 --------(1)-B+C+D+E=1 ---(2)-C+E=0 -------(3)

So, D=-B, E=C, and 2B+2C=1
Suppose B=0, then C=1/2, D=0, E=1/2
x/[(x+1)(x²-1)]=(1/2)/(x²-1)+(1/2)/(x+1)²

Suppose C=0, then B=1/2, D=-1/2, E=0
x/[(x+1)(x²-1)]=(x/2)/(x²-1)-(x/2)/(x+1)²

Hope this can help you. (though it is not the best method)

2012-01-15 16:50:34 補充：
數無盡 學無涯 : But why can't it work

Ans : the denominator is wrong, so it can't work. for example
13/24 = A/4 + B/6
you can't find any integers that suitable for A and B and the LCM of 4 and 6 is not 24.

Because the sixth line is not identity . So you cannot compare coefficient .

Notice that x^2 - 1 can be factored as (x - 1)(x + 1) and so it cannot be used in partial fraction directly.

In fact, x/[(x+1)(x²-1)]=A/(x+1)+(Bx+C)/(x²-1)

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

Hope it help