求不定积分∫1/(x^4*√(1+x^2))dx
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令x=tanθ,则:
√(1+x^2)=√[1+(tanθ)^2]=1/cosθ,
sinθ=√{(sinθ)^2/[(sinθ)^2+(cosθ)^2]}=√{(tanθ)^2/[(tanθ)^2+1]}
=tanθ/√[(tanθ)^2+1]=x/√(x^2+1),
dx=[1/(cosθ)^2]dθ。
∴原式=∫{1/[(tanθ)^4(1/cosθ)]}[1/(cosθ)^2]dθ
=∫{1/[(tanθ)^4cosθ]}dθ
=∫[(cosθ)^3/(sinθ)^4]dθ
=∫[(cosθ)^2/(sinθ)^4]d(sinθ)
=∫{[1-(sinθ)^2]/(sinθ)^4}d(sinθ)
=∫[1/(sinθ)^4]d(sinθ)-∫[1/(sinθ)^2]d(sinθ)
=-(1/3)[1/(sinθ)^3]+(1/sinθ)+C
=1/[x/√(x^2+1)]-(1/3)/[x/√(x^2+1)]^3+C
=√(x^2+1)/x-(x^2+1)√(x^2+1)/(3x^3)+C
√(1+x^2)=√[1+(tanθ)^2]=1/cosθ,
sinθ=√{(sinθ)^2/[(sinθ)^2+(cosθ)^2]}=√{(tanθ)^2/[(tanθ)^2+1]}
=tanθ/√[(tanθ)^2+1]=x/√(x^2+1),
dx=[1/(cosθ)^2]dθ。
∴原式=∫{1/[(tanθ)^4(1/cosθ)]}[1/(cosθ)^2]dθ
=∫{1/[(tanθ)^4cosθ]}dθ
=∫[(cosθ)^3/(sinθ)^4]dθ
=∫[(cosθ)^2/(sinθ)^4]d(sinθ)
=∫{[1-(sinθ)^2]/(sinθ)^4}d(sinθ)
=∫[1/(sinθ)^4]d(sinθ)-∫[1/(sinθ)^2]d(sinθ)
=-(1/3)[1/(sinθ)^3]+(1/sinθ)+C
=1/[x/√(x^2+1)]-(1/3)/[x/√(x^2+1)]^3+C
=√(x^2+1)/x-(x^2+1)√(x^2+1)/(3x^3)+C
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