If $f\left(x\right)=\int \frac{5x^8+7x^6}{{\left(x^2+2x^7+1\right)}^2}dx(x\ge 0)$ and $f\left(0\right)=0$, then the value of f(1)=

$f\left(x\right)=\int \frac{5x^8+7x^6}{{\left(x^2+2x^7+1\right)}^2}dx$

$f\left(x\right)=\int \frac{5x^8+7x^6}{{\left(x^7\right)}^2{\left(x^{-5}+2+x^{-7}\right)}^2}dx$

$f\left(x\right)=\int \frac{5x^8+7x^6}{x^{14}{\left(x^{-5}+2+x^{-7}\right)}^2}dx$

$f\left(x\right)=\int \frac{5x^{-6}+7x^{-8}}{{\left(x^{-5}+2+x^{-7}\right)}^2}dx$

Put

$x^{-5}+x^{-7}+2=t$

$-\left(5x^{-6}+7x^{-8}\right)dx=dt$

$=\int \frac{-1}{{\left(t\right)}^2}dx$

$f\left(x\right)=\frac{x^7}{\left(x^2+2x^7+1\right)}+c$

$f\left(0\right)=0$ then c=0

f(1)=$\frac{1}{4}$