∫dxx2+1x2+4=13∫3x2+1x2+4dx=13∫x2+4−x2+1x2+1x2+4dx=13∫1x2+1−1x2+4dx=13tan−1⁡x−12tan−1⁡x2+C=13tan−1⁡x−16tan−1⁡x2+C

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

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$\frac{1}{3} \tan^{- 1} x + \frac{1}{6} \tan^{- 1} (\frac{x}{2}) + C$

$\frac{1}{3} \tan^{- 1} x - \frac{1}{3} \tan^{- 1} (\frac{x}{2}) + C$

$\frac{1}{3} \tan^{- 1} x + \frac{1}{3} \tan^{- 1} (\frac{x}{2}) + C$

$\frac{1}{3} \tan^{- 1} x - \frac{1}{6} \tan^{- 1} (\frac{x}{2}) + C$

answer is D.

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Detailed Solution

$\begin{array}{l} \int \frac{d x}{(x^{2} + 1) (x^{2} + 4)} = \frac{1}{3} \int \frac{3}{(x^{2} + 1) (x^{2} + 4)} d x \\ = \frac{1}{3} \int \frac{(x^{2} + 4) - (x^{2} + 1)}{(x^{2} + 1) (x^{2} + 4)} d x \\ = \frac{1}{3} \int [\frac{1}{x^{2} + 1} - \frac{1}{x^{2} + 4}] d x \\ = \frac{1}{3} (\tan^{- 1} x - \frac{1}{2} \tan^{- 1} \frac{x}{2}) + C \\ = \frac{1}{3} \tan^{- 1} x - \frac{1}{6} \tan^{- 1} (\frac{x}{2}) + C \end{array}$