求不定积分∫(x^3)/(1+x^2)^3/2dx

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求不定积分∫(x^3)/(1+x^2)^3/2dx求不定积分∫(x^3)/(1+x^2)^3/2dx求不定积分∫(x^3)/(1+x^2)^3/2dx∵∫[x^3/(1+x^2)^(3/2)]dx=(

求不定积分∫(x^3)/(1+x^2)^3/2dx
求不定积分∫(x^3)/(1+x^2)^3/2dx

求不定积分∫(x^3)/(1+x^2)^3/2dx
∵∫[x^3/(1+x^2)^(3/2)]dx=(1/2)∫[(1+x^2-1)/(1+x^2)^(3/2)]d(1+x^2).
∴可令√(1+x^2)=y,得:1+x^2=y^2,∴d(1+x^2)=2ydy.
于是:
∫[x^3/(1+x^2)^(3/2)]dx
=(1/2)∫[(y^2-1)/y^3](2y)dy
=∫[(y^2-1)/y^2]dy
=∫dy-∫(1/y^2)dy
=y+1/y+C
=√(1+x^2)+1/√(1+x^2)+C.