Let $$I_n=\int \frac{(x^2-1)^n}{x(x^n)}\,dx.$$ , where x≠0
Then we define $$f(n,x)=nI_n + 4(n-1)I_{n-2}$$ If f(n,1)=0, what is $f(2,2)$?
I started with using integration by parts as it appears like a typical question where rewriting the fraction as a product of two fractions with powers $2$, $n-2$ helps but it ended up messy, is there a better way to approach or a better continuation to the typical method?
PS-the n in In is a subscript(would be great if anybody can edit it, couldn't locate the subscript option)
Note that $$f’(2,x)= 2 \frac{(x^2-1)^2}{x(x^2)}+4(2-1) \frac{1}{x} =\frac2{x^3}+2x $$ and $$f(2,2)=f(2,1)+\int_1^2 f’(2,x)\ dx=\frac{15}4 $$