Question

The value of $\int \frac{x^2-1}{(x^2+1)\sqrt{1+x^4}}dx$ is (where $C$ is integration constant)

• A. ${\sec }^{-1}\left(\frac{\sqrt{x^2+\frac{1}{x^2}}}{\sqrt{2}}\right)+C$
• B. $\frac{1}{2\sqrt{2}}{\sec }^{-1}\left(\frac{\sqrt{x^2+\frac{1}{x^2}}}{\sqrt{2}}\right)+C$
• C. $\frac{1}{\sqrt{2}}{\sec }^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)+C$
• D. $\frac{1}{2}{\sec }^{-1}\left(\frac{\sqrt{x^2+\frac{1}{x^2}}}{\sqrt{2}}\right)+C$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}{\sec }^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)+C$

Given,
$I=\int \frac{x^2-1}{(x^2+1)\sqrt{1+x^4}}dx=\int \frac{x^2(1-\frac{1}{x^2})dx}{x(x+\frac{1}{x})\times x\sqrt{x^2+\frac{1}{x^2}}}=\int \frac{(1-\frac{1}{x^2})dx}{(x+\frac{1}{x})\sqrt{{(x+\frac{1}{x})}^2-2}}$
Putting,
$x+\frac{1}{x}=y\Rightarrow (1-\frac{1}{x^2})dx=dy\Rightarrow I=\int \frac{dy}{y\sqrt{y^2-2}}=\frac{1}{\sqrt{2}}{\sec }^{-1}\frac{y}{\sqrt{2}}+C=\frac{1}{\sqrt{2}}{\sec }^{-1}\left(\frac{x+\frac{1}{x}}{\sqrt{2}}\right)+C$