- #1
dyn
- 773
- 62
- 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
$$ \int x/(2x-1) dx = \int x/[2(x-(1/2)] dx = (1/2) \int x/(x-(1/2)) dx $$ $$= (1/2) \int (x-(1/2)+(1/2))/(x-(1/2)) dx = (1/2) \int 1+(1/2)/(x-(1/2))dx = x/2+(1/4)ln|x-(1/2)|+C$$
What is wrong with my method ? I just can't see it.
Thanks