∫(dx/((1+x^1/3)x^1/2))计算不定积分

题目

∫(dx/((1+x^1/3)x^1/2))计算不定积分

答案

∫ 1/[(1+x^1/3)x^1/2] dx
令x^1/6=u,则x^1/2=u^3,x^1/3=u^2,x=u^6,dx=6u^5du
=∫ 6u^5/[(1+u^2)u^3] du
=6∫ u^2/(1+u^2) du
=6∫ (u^2+1-1)/(1+u^2) du
=6∫ (u^2+1)/(1+u^2) du - 6∫ 1/(1+u^2) du
=6u - 6arctanu + C
=6x^(1/6) - 6arctan[x^(1/6)] + C
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