Discussion Forum



1)

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


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

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

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

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

Answer:

Option C

Explanation:

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   $x-\frac{1}{x}=t$

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

 $\therefore$    

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

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