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x^1/2dx/(1+x^1/3)
题目详情
x^1/2dx/(1+x^1/3)
▼优质解答
答案和解析
求不定积分∫[(√x)/(1+∛x)]dx
令x¹/⁶=u,则x=u⁶;dx=6u⁵dx;代入原式得:
原式=6∫[u⁸/(1+u²)]du=6∫[u⁶-u⁴+u²-1+1/(1+u²)]du
=6(u⁷/7-u⁵/5+u³/3-u+arctanu)+C
=6[(1/7)x^(7/6)-(1/5)x^(5/6)+(1/3)x^(1/2)-x^(1/6)]+6arctan[x^(1/6)]+C
令x¹/⁶=u,则x=u⁶;dx=6u⁵dx;代入原式得:
原式=6∫[u⁸/(1+u²)]du=6∫[u⁶-u⁴+u²-1+1/(1+u²)]du
=6(u⁷/7-u⁵/5+u³/3-u+arctanu)+C
=6[(1/7)x^(7/6)-(1/5)x^(5/6)+(1/3)x^(1/2)-x^(1/6)]+6arctan[x^(1/6)]+C
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