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partial fractions.. x/x^2-4
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Write it as \[ \frac{x}{x^2-4}=\frac{A}{x-2}+\frac{B}{x+2} \]
Find A and B
Multiply the two sides by (x-2) to get \[ \frac{x (x-2)}{(x-2) (x+2)}=\frac{A (x-2)}{x-2}+\frac{B (x-2)}{x+2}\\ \frac{x}{x+2}=\frac{A}{1}+\frac{B (x-2)}{x+2} \] Put x =2 and get A Do the same for B
thank you
The final answer looks like \[ \frac{x}{x^2-4}=\frac{1}{2 (x+2)}+\frac{1}{2 (x-2)} \]
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YW
thats the final answer
yes
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