∫x4−x2+1+x21+x6dx∫x4−x2+1dx13+x23+x21+x32dx∫x4−x2+11+x21−x2+x4dx+13∫3x21+x32dx=Tan−1x+Tan−1x3+c

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$\int \frac{x^{4} + 1}{x^{6} + 1} d x =$

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$\tan^{- 1} x - \tan^{- 1} x^{3} + c$

$\tan^{- 1} x - \frac{1}{3} t a n^{- 1} (x^{3}) + c$

$\tan^{- 1} x + \tan^{- 1} (x^{3}) + c$

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

answer is D.

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Detailed Solution

$\int \frac{(x^{4} - x^{2} + 1) + x^{2}}{1 + x^{6}} d x$ $\int \frac{x^{4} - x^{2} + 1 d x}{1^{3} + {(x^{2})}^{3}} + \frac{x^{2}}{1 + {(x^{3})}^{2}} d x$

$\int \frac{(x^{4} - x^{2} + 1)}{(1 + x^{2}) (1 - x^{2} + x^{4})} d x + \frac{1}{3} \int \frac{3 x^{2}}{1 + {(x^{3})}^{2}} d x$ $= T a n^{- 1} x + T a n^{- 1} (x^{3}) + c$