求解求解数学
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解:∫[1/(x²-2x-3)]=∫{1/[(x+1)(x-3)]}dx=(1/4)∫[1/(x-3)-1/(x+1)]dx
=(1/4)[∫[dx/(x-3)-∫dx/(x+1)]
=(1/4)[ln(x-3)-ln(x+1)]+C
=[ln(x-3)-ln(x+1)]/4+C
=ln[(x-3)/(x+1)]/4+C
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看图 ̄  ̄)σ
解:∫[1/(x²-2x-3)]=∫{1/[(x+1)(x-3)]}dx=(1/4)∫[1/(x-3)-1/(x+1)]dx
=(1/4)[∫[dx/(x-3)-∫dx/(x+1)]
=(1/4)[ln(x-3)-ln(x+1)]+C
=[ln(x-3)-ln(x+1)]/4+C
=ln[(x-3)/(x+1)]/4+C
下一篇:一道题目求解,数学的,不要列方程
上一篇:太阳王()猜谜语