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