Simplify $logleft(1+frac{x_i^2}{nu}right)$ with a $log(1+x)$ rule?

There is no realy usefull simplification... But if was about approximation, you could use the fact that $$ log(1+x) = x- frac{x^2}{2} + O(x^3) $$

to say that, for small $x$, $$ log(1+frac{x_i^2}{nu}) approx frac{x_i^2}{nu}-frac 12 left(frac{x_i^2}{nu}right)^2 = frac{x_i^2}{nu}left(1- frac{x_i^2}{2 nu}right). $$

If the $1$ was not there, we could have written the expression as $$2log x_i-lognu$$ provided that the variables are positive. But the $1$ prevents this, though it does mean we get a Taylor series expansion $$logleft(1+frac{x_i^2}nuright)=sum_{n=1}^infty(-1)^{n+1}frac{(x_i^2/nu)^n}n$$