Rewrite the integrand by dividing numerator and denominator by $x^{18}$:

$f(x)=\int \frac{\frac{7}{x^8}+\frac{9}{x^{10}}}{{\left(\frac{1}{x^9}+\frac{1}{x^7}+2\right)}^2}dx$

Substitute $t=\frac{1}{x^9}+\frac{1}{x^7}+2$, so $dt=\left(-\frac{9}{x^{10}}-\frac{7}{x^8}\right)dx$. Then

$f(x)=-\int \frac{dt}{t^2}=\frac{1}{t}+C=\frac{x^9}{1+x^2+2x^9}+C$

$f(1)=\frac{1}{4}+C=\frac{1}{4}\Rightarrow C=0$, consistent with ${\lim }_{x\to 0}f(x)=0$.

Differentiate: $f^{\prime }(x)=\frac{9x^8(1+x^2+2x^9)-x^9(2x+18x^8)}{(1+x^2+2x^9{)}^2}$. At $x=1$: $f^{\prime }(1)=\frac{9\cdot 4-1\cdot 20}{16}=\frac{16}{16}=1$.

So $A=\left[\begin{array}{c}001\\ 1/411\\ {\alpha }^{2}41\end{array}\right]$. Expanding along row 1: $|A|=1\left(\frac{1}{4}\cdot 4-1\cdot {\alpha }^2\right)=1-{\alpha }^2$.

For a $3\times 3$ matrix, $|\text{adj}(\text{adj}A)|=|A{|}^{(n-1{)}^2}=|A{|}^4=81$, so $|A|=\pm 3$.

$1-{\alpha }^2=\pm 3\Rightarrow {\alpha }^2=-2$ (rejected) or ${\alpha }^2=4$.