Solving this integral in 2 different ways

[dyn]

Homework Statement

integrate x/(2x-1) with respect to x

Relevant Equations

$\int \frac 1 x \, dx = \ln|x|$

The answer gives $$ \int x /(2x-1)\ dx = x/2 +(1/4)ln|2x-1| + C $$ whicjh I can obtain. But when I try a different way I get a different answer. I must be making a stupid mistake but I can't see it. Here is my method
$$ \int x/(2x-1) dx = \int x/[2(x-(1/2)] dx = (1/2) \int x/(x-(1/2)) dx $$ $$= (1/2) \int (x-(1/2)+(1/2))/(x-(1/2)) dx = (1/2) \int 1+(1/2)/(x-(1/2))dx = x/2+(1/4)ln|x-(1/2)|+C$$
What is wrong with my method ? I just can't see it.
Thanks

 

[Charles Link]

$\ln|x-\frac{1}{2}|=\ln|(2x-1)/2|=\ln|2x-1|-\ln|2| =\ln|2x-1|+C'$. What you did is equally correct.

 

[dyn]

Thank you. That problem has been driving me crazy!