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求不定积分∫(x^1/2)/(1-x^1/3)dx
人气:153 ℃ 时间:2020-01-30 14:35:46
解答
令u=x^(1/6),x=u^6,dx=6u^5 du∴∫√x/(1-x^1/3) dx=6∫u^8/(1-u²) du用长除法:u^8=(1-u²)(-u^6-u⁴-u²-1)+1u^8/(1-u²)=-u^6-u⁴-u²-1+1/(1-u²)∴6∫u^8/(1-u²) du=6...
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