Question

dx/x平方(1-x)的不定积分是多少

Answer 1

1/[x^2(1-x)]=A/(x-1)+B/x+C/x^2

Ax^2+Bx(x-1)+C(x-1)=-1

x=0:C=1 x=1:A=-1 x=-1:A+2B-2C=-1 B=1

∫dx/[x^2(1-x)]=-∫dx/(x-1)+∫dx/x+∫dx/x^2

=-ln|x-1|+ln|x|-1/x+C

=ln|x/(x-1)|-1/x+C