$\text{ If }\int \frac{x^2-4}{x^4+9x^2+16}dx=A{\tan }^{-1}\left(f\right(x\left)\right)+B,\text{ then the value of }f\left(10\right)\text{ is }$

$I=\int \frac{x^2-4}{x^4+9x^2+16}dx$

Divide both numerator and denominator by x²
$\begin{array}{l}\mathrm{I}=\int \frac{\left(1-\frac{4}{{\mathrm{x}}^2}\right)}{{\mathrm{x}}^2+9+\frac{16}{{\mathrm{x}}^2}}\mathrm{dx}=\int \frac{\left(1-\frac{4}{{\mathrm{x}}^2}\right)}{{\left(\mathrm{x}+\frac{4}{\mathrm{x}}\right)}^2+1}\mathrm{dx}\\ \text{ Put }\mathrm{u}=\mathrm{x}+\frac{4}{\mathrm{x}},\mathrm{du}=\left(1-\frac{4}{{\mathrm{x}}^2}\right)\mathrm{dx}\\ \mathrm{I}=\int \frac{\mathrm{du}}{{\mathrm{u}}^2+1}={\tan }^{-1}\left(\mathrm{u}\right)+\mathrm{B}\\ =1\cdot {\tan }^{-1}\left(\mathrm{x}+\frac{4}{\mathrm{x}}\right)+\mathrm{B}\\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+\frac{4}{\mathrm{x}}\end{array}$