Integrating $\displaystyle\int_2^3 \dfrac{\text{d}x}{2x(1-2x)}$ in W. Mathematica giving wrong result. I calculated by hand the following integral
$$\int_2^3 \dfrac{\text{d}x}{2x(1-2x)}$$
What I did it to use partial fractions:
$$\dfrac{1}{x(1-2x)} = \dfrac{1}{x} + \dfrac{-2}{1-2x}$$
Whence
$$\dfrac{1}{2}\int_2^3 \left(\dfrac{1}{x} + \dfrac{-2}{1-2x}\right) = \dfrac{1}{2}\left(\ln|x|\big|_2^3 + \ln|1-2x|\big|_2^3\right) = \dfrac{1}{2}\ln\left(\dfrac{5}{2}\right)$$
Now, checking with W. Mathematica (NOT W. Alpha) it does return the value of $\ln\dfrac{3}{\sqrt{10}}$
How is it possible? Even with some manipulation, the max I can get is $\ln\dfrac{\sqrt{10}}{2}$... Thank you!

The partial fraction that you got is wrong.
$$\frac{1}{x(1-2x)} \ne \frac{1}{x} + \frac{-2}{1-2x}$$
Since,
$$\frac{1}{x} + \frac{-2}{1-2x} = \frac{(1-2x)-2x}{x(1-2x)} = \frac{1-4x}{x(1-2x)}$$
