If $\int \frac{x^2+4}{x^4+16} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x^2-4}{a x}\right)+C$, then $a=$
Answer & explanation
Correct answer: option 2
Let
$I=\int \frac{x^2+4}{x^4+16} d x=\int \frac{1+\frac{4}{x^2}}{x^2+\frac{16}{x^2}-8+8} d x$
$\Rightarrow I=\int \frac{1}{\left(x-\frac{4}{x}\right)^2+(2 \sqrt{2})^2} d\left(x-\frac{4}{x}\right)$
$\Rightarrow I=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left\{\frac{x-\frac{4}{x}}{2 \sqrt{2}}\right\}+C=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left\{\frac{x^2-4}{2 \sqrt{2} x}\right\}+C$
Hence, $a=2 \sqrt{2}$.