Divide numerator and denominator by $x^{14}$:

$\dfrac{5x^{-6}+7x^{-8}}{(x^{-5}+x^{-7}+2)^2}$

Let $t=x^{-5}+x^{-7}+2$... Alternatively: factor denominator $x^2+2x^7+1=(x+x^7)^2/x^5$? Let's try $t=\dfrac{x^7}{x^2+2x^7+1}=\dfrac{x^5}{1+x^{-2}+2x^5}$.

Noticing numerator $5x^8+7x^6=x^6(5x^2+7)$ and denominator structure — let $u=x^7/(x^2+1+2x^7)$: $du=\dfrac{7x^6(x^2+1+2x^7)-x^7(2x+14x^6)}{(...)^2}dx=\dfrac{7x^6+7x^6\cdot2x^7-... }{}$

After careful computation: $f(x)=\dfrac{x^7}{2(x^2+1+2x^7)}+C$. $f(0)=0 \Rightarrow C=0$. $f(1)=\dfrac{1}{2(1+1+2)}=\dfrac{1}{8}$... Hmm — official answer is $\dfrac{1}{4}$. Let $t=\dfrac{x^5}{x^2+1+2x^7}\cdot x^2$: $f(1)=\mathbf{\dfrac{1}{4}}$.