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1/x^2+4x+3+ 1/ x^2-1 ...please help me solve this
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is it this\[\huge \frac{1}{x^2} +4x + 3 + \frac{1}{x^2-1}\]?
no the 4x+3 goes under 1 in the first fraction
\[\frac{1}{x^2+4x+3}+ \frac{1}{x^2-1}\]right?
yes :)
let's see what i can do with this, just simplify i guess first factor \[x^2 + 4x + 3 \space and \space x^2 - 1\]
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(x+3) (x+1) and (x+1) (x-1)
\[\frac{1}{(x+3)(x+1)} + \frac{1}{(x+1)(x-1)}\]make the denominators common~
that's what I don't get..I'm stuck there
\[\frac{(x-1)+(x+3)}{(x+3)(x+1)(x-1)}\]rings any bells? :s
sort of
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