Can someone please show that
$\displaystyle \frac{1}{(x + 2)^2 (x - 1)} = \frac{A}{x + 2} + \frac{B}{(x + 2)^2} + \frac{C}{x - 1}$
is = -1/9/x + 2 - 1/3 / (x + 2)^2 + 1/9 / x - 1
Can someone please show that
$\displaystyle \frac{1}{(x + 2)^2 (x - 1)} = \frac{A}{x + 2} + \frac{B}{(x + 2)^2} + \frac{C}{x - 1}$
is = -1/9/x + 2 - 1/3 / (x + 2)^2 + 1/9 / x - 1
Standard method: multiply both sides by that denominator, $\displaystyle (x+2)^2(x-1)$.
1= A(x+2)(x-1)+ B(x-1)+ C(x+2)^2.
Now, letting x= -2 give B immediately. Letting x= 1 gives an equation involving both A and C but then letting x= 0 gives another equation in A and C and you can solve those two equations.
Asked here: http://www.mathhelpforum.com/math-he...ns-117360.html
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