\(\int \frac{x^2-4}{x^4+9 x^2+16} \cdot \mathrm{~d} x=\tan ^{-1}(\mathrm{f}(x))+\mathrm{c}(\) where c is a constant of integration), then value of \(f(2)\) is

Let \(I=\int \frac{x^2-4}{x^4+9 x^2+16} \mathrm{~d} x\)
\(\begin{aligned} & =\int \frac{1-\frac{4}{x^2}}{x^2+\frac{16}{x^2}+9} d x \\ & =\int \frac{1-\frac{4}{x^2}}{\left(x+\frac{4}{x}\right)^2+1} d x\end{aligned}\)
Put \(x+\frac{4}{x}=\mathrm{t} \Rightarrow\left(1-\frac{4}{x^2}\right) \mathrm{d} x=\mathrm{dt}\)
\(\begin{aligned} \therefore \quad \mathrm{I} & =\int \frac{\mathrm{dt}}{\mathrm{t}^2+1} \\ & =\tan ^{-1} \mathrm{t}+\mathrm{c} \\ & =\tan ^{-1}\left(x+\frac{4}{x}\right)+\mathrm{c}\end{aligned}\)
\(\begin{array}{ll}\therefore \quad & \mathrm{f}(x)=x+\frac{4}{x} \\ & \Rightarrow \mathrm{f}(2)=2+2=4\end{array}\)