Find $\int\frac{dx}{2x^2-2x+1}$

We can solve this by rewriting the function and completing the square 
\[ \int\frac{dx}{2x^2-2x+1}=\int \frac{dx}{2(x^2-x+\frac{1}{4})+\frac{1}{2}}=\int \frac{dx}{2(x-\frac{1}{2})^2+\frac{1}{2}}\]
\[=\frac{1}{2}\int \frac{dx}{(x-\frac{1}{2})^2+\frac{1}{4}}=\frac{1}{2}\arctan (\frac{x-\frac{1}{2}}{\frac{1}{2}})+c=\frac{1}{2}\times \frac{1}{2}\arctan (2x-1)+c\]
\[=\frac{1}{4}\arctan (2x-1)+c.\]
Here we have used the antiderivative formula
\[\int \frac{dx}{(x-b)^2+a^2}=\frac{1}{a} \arctan (\frac{x-a}{a})+c.\]