Q. The value of $\int_{}^{}{\frac{x^{2} + 1}{x^{4} - x^{2} + 1}dx}$ is
A) $\tan^{- 1}{\left( 2x^{2} - 1 \right) + c}$
B) $\tan^{- 1}{\frac{x^{2} + 1}{x} + c}$
C) $\sin^{- 1}{\left( x - \frac{1}{x} \right) + c}$
D) $\tan^{- 1}{\left( \frac{x^{2} - 1}{x} \right) + c}$
Explanation
(d)
Let $I = \int_{}^{}{\frac{x^{2} + 1}{x^{4} - x^{2} + 1}dx = \int_{}^{}\frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} - 1}}dx$
$= \frac{\left( 1 + \frac{1}{x^{2}} \right)}{\left( x - \frac{1}{x} \right)^{2} + 1}dx$
Put $x - \frac{1}{x} = t\ \ \Rightarrow \ \ \left( 1 + \frac{1}{x^{2}} \right)dx = dt$
$\therefore\ I = \int_{}^{}\frac{dt}{t^{2} + 1} = \tan^{- 1}{t + c}$
$= \tan^{- 1}{\left( x - \frac{1}{x} \right) + c}$
$= \tan^{- 1}\left( \frac{x^{2} - 1}{x} \right) + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.