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$\int\frac{x^{2}+1}{x^{4}-x^{2}+1}dx=$ .......

$\tan^{-1}\left(\frac{x^{2}+1}{2}\right)+c$

$\tan^{-1}(x^{2})+c$

$\tan^{-1}\left(\frac{x^{2}-1}{2}\right)+c$

$\tan^{-1}(2x^{2}-1)+c$

Option C

Let l= $\int\frac{x^{2}+1}{x^{4}-x^{2}+1}dx$

= $\int\frac{1+\frac{1}{x^{2}}}{x^{2}-1+\frac{1}{x^{2}}}dx=\int\frac{1+\frac{1}{x^{2}}}{(x-\frac{1}{x})^{2}+1}$

put $x-\frac{1}{x}=t$

$(1+\frac{1}{x^{2}})dx=dt$

$\therefore$

$l=\int \frac{dt}{t^{2}+1})= \tan^{-1}(t)+c$

$= \tan^{-1}(x-\frac{1}{x})+c= \tan^{-1}(\frac{x^{2}-1}{x})+c$

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