求一个不定积分的题目，谢谢
\u6c42\u4e00\u4e2a\u4e0d\u5b9a\u79ef\u5206\u7684\u9898\u76ee\uff0c\u8c22\u8c22\u7b54\uff1a
\u222b[1/³\u221a(5-3x)]dx
=(-1/3)* \u222b [(5-3x)^(-1/3)] d(5-3x)
=(-1/3)*(3/2)*(5-3x)^(-1/3+1)+C
=-(1/2)*(5-3x)^(2/3)+C
\u5982\u56fe\uff1a
dx=sect·tantdt
代入原式，得
原式=∫tant/sect ·secttantdt
=∫tan²tdt
=∫(sec²t-1)dt
=∫sec²tdt-∫dt
=tant-t+c
cost=1/x
t=arccos1/x
tant=√1-x²
所以
原式=√1-x² -arccos1/x+c
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