Question

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

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

Answer 1

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

$I=\int \frac{x^2-1}{(x^2+1)\sqrt{1+x^4}}dx=\int \frac{(1-\frac{1}{x^2})x^2}{x(x+\frac{1}{x})x\sqrt{x^2+\frac{1}{x^2}}}dx=\int \frac{(1-\frac{1}{x^2})}{(x+\frac{1}{x})\sqrt{{(x+\frac{1}{x})}^2-2}}dx$
Substituting $(x+\frac{1}{x})=t\Rightarrow (1-\frac{1}{x^2})dx=dt$
We get
$I=\int \frac{dt}{t\sqrt{t^2-2}}=\int \frac{dt}{t^2\sqrt{1-\frac{2}{t^2}}}$
Again substituting
$\frac{\sqrt{2}}{t}=4\Rightarrow \frac{\sqrt{2}}{t^2}dt=du$
We get
$I=-\frac{1}{\sqrt{2}}\int \frac{du}{\sqrt{1-u^2}}=\frac{1}{\sqrt{2}}{\cos }^{-1}u+c=\frac{1}{\sqrt{2}}{\cos }^{-1}\left(\frac{\sqrt{2}}{x+\frac{1}{x}}\right)+c=\frac{1}{\sqrt{2}}{\cos }^{-1}\left(\frac{\sqrt{2}x}{x^2+1}\right)+c=\frac{1}{\sqrt{2}}{\sec }^{-1}\left(\frac{x^2+1}{\sqrt{2}x}\right)+c$