∫[1/(3+2x-x²)]dx = ∫dx/[(1+x)(3-x)] = (1/4)∫[1/(1+x) + 1/(3-x)]dx
= (1/4)[ln|1+x| - ln|3-x|] + C = (1/4)ln|(1+x)/(3-x)| + C
∫1/(3+2x-x²)]dx
= ∫dx/[(1+x)(3-x)]
=¼∫[1/(1+x) + 1/(3-x)]dx
=¼[ln|1+x| - ln|3-x|] + C
=¼ln|(1+x)/(3-x)| + C
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