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
=a/(x+1)+b/(x-1)
=[(a+b)x+(b-a)]/(x+1)(x-1)
所以a+b=0,b-a=1
a=-1/2,b=1/2
所以原式=-1/2∫1/(x+1)dx+1/2∫1/(x-1)dx
=-1/2*ln|x+1|+1/2*ln|x-1|+C
=1/2*ln|(x-1)/(x+1)|+C
你把x2-1分解公因式，得到（x-1）*（x+1），然后1/（x2-1）=1/2（x-1) - 1/2（x+1)
剩下的套对数公式就行了
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绛旓細绛旓細鈭玔1/³鈭(5-3x)]dx =(-1/3)* 鈭 [(5-3x)^(-1/3)] d(5-3x)=(-1/3)*(3/2)*(5-3x)^(-1/3+1)+C =-(1/2)*(5-3x)^(2/3)+C
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