∫1/[（x^1/3）+（x^1/2）]dx求不定积分

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∫1/[（x^1/3）+（x^1/2）]dx求不定积分∫1/[（x^1/3）+（x^1/2）]dx求不定积分∫1/[（x^1/3）+（x^1/2）]dx求不定积分令x^(1/6)=t,则x^(1/3)

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

∫1/[（x^1/3）+（x^1/2）]dx求不定积分
令x^(1/6)=t,则x^(1/3)=t^2,x^(1/2)=t^3,x=t^6,dx=6t^5dt
于是,原式=∫6t^5dt/(t^2+t^3)
=6∫t^3dt/(t+1)
=6∫[t^2-t+1-1/(t+1)]dt
=6(t^3/3-t^2/2+t-ln│t+1│)+C (C是常数)
=2t^3-3t^2+6t-6ln│t+1│+C
=2x^(1/2)-3x^(1/3)+6x^(1/6)-6ln│x^(1/6)+1│+C.

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