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求x的三次方dx 除以(1-x^2)^3÷2的不定积分
人气:377 ℃ 时间:2020-04-10 09:42:18
解答
∫[x^3/(1-x^2)^(3/2)]dx
=(1/2)∫[x^2/(1-x^2)^(3/2)]d(x^2)
=-(1/2)∫[(1-1+x^2)/(1-x^2)^(3/2)]d(1-x^2)
=-(1/2)∫[1/√(1-x^2)^3]d(1-x^2)+∫[1/√(1-x^2)]d(1-x^2)
=-(1/2)∫(1-x^2)^(-3/2)d(1-x^2)+(1/2)∫(1-x^2)^(-1/2)d(1-x^2)
=-(1/2)×(-2)(1-x^2)^(-1/2)+(1/2)×2(1-x^2)^(1/2)+C
=1/√(1-x^2)+√(1-x^2)+C
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