Question

1/ 2x(x-1) dx 这个怎么积分。 要详细过程。。。。。

Answer 1

1/ 2x(x-1)=[1/2(x-1)]-1/2x,又因∫1/2(x-1)dx=1/2∫1/(x-1)dx=1/2∫1/(x-1)d（x-1）=[ln(x-1)]/2+c，∫1/2xdx=1/2∫1/xdx=[lnx]/2+c. 因此∫1/ 2x(x-1) dx=∫1/2(x-1)dx-∫1/2xdx=[ln(x-1)]/2-[lnx]/2+c也可化成[ln(x-1)/x]/2+c

Answer 2

1/2x(x-1)=1/2{1/(x-1)-1/x},所以积分得1/2{ln|x-1|-ln|x|}+C=1/2ln|1-1/x|+C

Answer 3

原式=∫1/2(x-1)-1/2xdx=1/2(∫1/(x-1)dx-∫1/xdx)=1/2ln(x-1/x)+c

Answer 4

设
A/2x+B/(x-1)=[A(x-1)+B*2x]/2x(x-1)
=(Ax-A+2Bx)/2x(x-1)
=[(A+2B)x-A]/2x(x-1)
=1/2x(x-1)
则有
-A=1
A=-1
A+2B=0
2B=-A=1
B=1/2
所以
1/2x(x-1)=1/2(x-1)-1/2x
所以
∫dx/2x(x-1)
=∫dx/2(x-1)-∫dx/2x
=1/2∫d(x-1)/(x-1)-1/2∫dx/x
=1/2(ln|x-1|-ln|x|)
=1/2ln|(x-1)|/|x|