rearrange f(x) for easier integration

Question

given a probability density function: $$f(x)=\frac{10}{(\pi)(1+(x+2)^2)(1+(x-2)^2}$$

I've been ask to find the cumulative distribution function, I have also been given a hint : Show the integrand can be rearranged as: $$f(x) = (\frac{1}{4\pi})(\frac{x+2}{(1+(x+2)^2)}+\frac{2}{(1+(x+2)^2)}-\frac{x-2}{(1+(x-2)^2)}+\frac{2}{(1+(x-2)^2)})$$

I'm unable to show the original function can be rearranged to give the function in the hint, I have attempted using partial fractions but could not get the rearranged fuction.

Answer 1

I found it a simple exercise to show that the two integrands are equivalent by quite simple algebra. Be careful, very careful of your signs. Just re-combine the separate fractions to show that it gives the original integrand.

Answer 2

Hint

If you have no idea you can always go backward: \begin{align} &\frac{x+2}{(1+(x+2)^2)}+\frac{2}{(1+(x+2)^2)}-\frac{x-2}{(1+(x-2)^2)}+\frac{2}{(1+(x-2)^2)}=\cdots\\ &\cdots=\frac{(4+x)(1+(x-2)^2)+(4-x)(1+(x+2)^2)}{(1+(x+2)^2)(1+(x-2)^2)}=\frac{40}{(1+(x+2)^2)(1+(x-2)^2)} \end{align} as: $$(4+x)(1+(x-2)^2)+(4-x)(1+(x+2)^2)=40$$