Q. $\int_{}^{}\frac{x^{4} - 1}{x^{2}\left( x^{4} + x^{2} + 1 \right)^{1/2}}\ dx$ is equal to
A) $\sqrt{\frac{x^{4} + x^{2} + 1}{c}} + c$
B) $\frac{x^{2}}{\sqrt{x^{4} + x^{2} + 1}} + c$
C) $x\left( x^{4} + x^{2} + 1 \right)^{3/2} + c$
D) $\frac{\sqrt{x^{4} + x^{2} + 1}}{x} + c$
Explanation
(d)
$\int_{}^{}\frac{x^{4} - 1}{x^{2}\left( x^{4} + x^{2} + 1 \right)^{1/2}}\ dx$
$= \int_{}^{}\frac{2x^{4} + x^{2} - (x^{4} + x^{2} + 1)}{x^{2}\sqrt{x^{4} + x^{2} + 1}}\ dx$
$= \int_{}^{}\frac{\frac{x(4x^{3} + 2x)}{2\sqrt{x^{4} + x^{2} + 1}} - \sqrt{x^{4} + x^{2} + 1}}{x^{2}}\ dx$
$= \int_{}^{}{\frac{d}{dx}\left( \frac{\sqrt{x^{4} + x^{2} + 1}}{x} \right) = \frac{\sqrt{x^{4} + x^{2} + 1}}{x}} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.