求∫√xdx/(√x-³√x)=? 5
令x=t^6,然后搞出这个∫6t^6dt/(t-1)，再怎么弄？令t-1=s吗？有没有更简单的？...
令x=t^6,然后搞出这个∫6t^6dt/(t-1)，再怎么弄？令t-1=s吗？有没有更简单的？
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∫[√x/(√x-³√x)] dx
let
y = x^(1/6)
dy = (1/6)x^(-5/6) dx
dx = 6y^5 dy
∫[√x/(√x-³√x)] dx
=6∫[y^8/(y^3-y^2)] dy
=6∫[y^6/(y-1)] dy
=6∫( y^5+y^4+y^3+y^2+y+1) dy + 6∫dy/(y-1)
=6[ (1/6)y^6+(1/5)y^5+(1/4)y^4+(1/3)y^3+(1/2)y^2 +y + ln|y-1|) + C'
=6[ (1/6)x^6+(1/5)x^(5/6)+(1/4)x^(2/3)+(1/3)x^(1/2)+(1/2)x^(1/3) +x^(1/6) + ln|x^(1/6)-1| ] + C'
let
y = x^(1/6)
dy = (1/6)x^(-5/6) dx
dx = 6y^5 dy
∫[√x/(√x-³√x)] dx
=6∫[y^8/(y^3-y^2)] dy
=6∫[y^6/(y-1)] dy
=6∫( y^5+y^4+y^3+y^2+y+1) dy + 6∫dy/(y-1)
=6[ (1/6)y^6+(1/5)y^5+(1/4)y^4+(1/3)y^3+(1/2)y^2 +y + ln|y-1|) + C'
=6[ (1/6)x^6+(1/5)x^(5/6)+(1/4)x^(2/3)+(1/3)x^(1/2)+(1/2)x^(1/3) +x^(1/6) + ln|x^(1/6)-1| ] + C'
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