Question

$\int \frac{(x^2-1)}{(x^2+1)\sqrt{x^4+1}}dx$ is equal to

• A. ${\sec }^{-1}\left(\frac{x^2+1}{\sqrt{2}x}\right)+c$
• B. $\frac{1}{\sqrt{2}}{\sec }^{-1}\left(\frac{x^2+1}{\sqrt{2}x}\right)+c$
• C. $\frac{1}{\sqrt{2}}{\sec }^{-1}\left(\frac{x^2+1}{\sqrt{2}}\right)+c$
• D. None of these

Answer 1

Answer: (B)

$I=\int \frac{x^2\left(1-\frac{1}{x^2}\right)dx}{x^2\left(x+\frac{1}{x}\right){\left(x^2+\frac{1}{x^2}\right)}^{1/2}}$
Let $x+\frac{1}{x}=p$
$\Rightarrow \left(1-\frac{1}{x^2}\right)dx=dp$
$I=\int \frac{dp}{p\sqrt{p^2-2}}=\frac{1}{\sqrt{2}}{\sec }^{-1}\frac{p}{\sqrt{2}}$
$=\frac{1}{\sqrt{2}}{\sec }^{-1}\frac{p}{\sqrt{2}}$
$=\frac{1}{\sqrt{2}}{\sec }^{-1}\left(\frac{x^2+1}{\sqrt{2}x}\right)+c$