∫x2+4x4+16 dx=1Ktan−1x2−4Kx+C then K=

∫x2+4x4+16 dx=∫1+4x2x2+16x2 dx=∫1+4x2dxx−4x2+8=122tan−1x−4x22+C∵∫f1xfx2+a2 dx=1atan−1fxa

∫x2+4x4+16 dx=1Ktan−1x2−4Kx+C then K=

∫x2+4x4+16 dx=∫1+4x2x2+16x2 dx=∫1+4x2dxx−4x2+8=122tan−1x−4x22+C∵∫f1xfx2+a2 dx=1atan−1fxa

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$\int \frac{x^{2} + 4}{x^{4} + 16} d x = \frac{1}{K} \tan^{- 1} (\frac{x^{2} - 4}{K x}) + C$ then $K =$

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$\sqrt{2}$

$4 \sqrt{2}$

$2 \sqrt{2}$

answer is C.

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$\int \frac{x^{2} + 4}{x^{4} + 16} d x = \int \frac{1 + \frac{4}{x^{2}}}{x^{2} + \frac{16}{x^{2}}} d x = \int \frac{(1 + \frac{4}{x^{2}}) d x}{{(x - \frac{4}{x})}^{2} + 8}$

$= \frac{1}{2 \sqrt{2}} \tan^{- 1} (\frac{x - \frac{4}{x}}{2 \sqrt{2}}) + C$

$∵ \int \frac{f^{1} (x)}{f {(x)}^{2} + a^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{f (x)}{a})$

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$\int \frac{x^{2} + 4}{x^{4} + 16} d x = \frac{1}{K} \tan^{- 1} (\frac{x^{2} - 4}{K x}) + C$ then $K =$