∫1/x^2(1-x)dx求不定积分

Question

∫ 1/x^2(1-x)dx
求不定积分

Answer 1

答：
1/[(1-x)x^2)]=a/(1-x)+(bx+c)/x^2=(ax^2+bx+c-bx^2-cx)/[(1-x)x^2]
a-b=0
b-c=0
c=1
解得：a=b=c=1
∫{1/[(1-x)x^2]}dx
=∫[1/(1-x)]dx+∫[(x+1)/x^2]dx
=-ln|1-x|+ln|x|-1/x+C
=ln|x/(1-x)|-1/x+C