∫(1,2)x∧2–2x–3/xdx定积分 - 知识问答

∫(1,2)x∧2–2x–3/xdx定积分

1个回答

x/(x²-2x-3)
=(x-3+3)/(x-3)(x+1)
=1/(x+1)+3/(x-3)(x+1)
=1/(x+1)+3/4*[1/(x-3)-1/(x+1)]
=3/4*1/(x-3)+1/4*1/(x+1)
所以原式=3/4*ln|x-3|+1/4*ln|x+1|+C

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