# Comparing fractions by using partial fractions or a specific factorisation?

Trying to reproduce the results of a paper I had to solve a difficult integral. I obtain:

$$$$A=-N\left ( \epsilon r+\log(1+r) +\frac{rx^2(1-r\tau + \tau r^2 (\tau +1))}{1+r}+\frac{ry^2(-1+r\tau + \tau r^2 (\tau -1))}{1+r}\right)$$$$

However the correct result is: $$$$B=-N\left ( \epsilon r+\log(1+r) +\frac{rx^2}{1+r(1+\tau)}+\frac{ry^2}{1+r(1-\tau)}\right)$$$$

I am convinced that the factor $$(1+r)$$ is correct since it is present in the $$\log$$. However I do not know how the authors came up with their denominators.

Comparing $$A$$ and $$B$$ with specific numerical values shows they are not the same. But when I do $$B-A$$ on mathematica it does not provide me any help.

Ideally I would like to put the fractions of $$B$$ in the following form:

$$\frac{a_1}{1+r}+\frac{a_2}{1+r}+\frac{a_3}{1+r}+\dots$$ Then, I could immediately spot which terms are wrong in my answer.

Would you know how to do this? I am open for any advice or different idea, thank you very much.

• Please help us help you by entering your equations in Mathematica code with proper syntax, properly formatted in code blocks. (Click the grey edit button below your post to edit, and see the help for how to type equations in code blocks.) That way, we can copy and paste your code into our own copies of Mathematica without having to take the time to type all that out ourselves. – march Feb 11 at 16:51