Question

$\int \frac{x^2-1}{(x^2+1)\sqrt{1+x^4}}dx=k{\cos }^{-1}\left(\frac{\sqrt{2}x}{x^2+1}\right)+C$, where $k=$

• A. $\frac{1}{2}$
• B. $2$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}$

Let $I=\int \frac{x^2-1}{(x^2+1)\sqrt{1+x^4}}dx=\int \frac{(1+\frac{1}{x^2})x^2}{x(1-\frac{1}{x})x\sqrt{x^2+\frac{1}{x^2}}}dx$
$=\int \frac{(1-\frac{1}{x^2})x^2}{(1+\frac{1}{x})\sqrt{{(x+\frac{1}{x})}^2-2}}dx$
Substitute $x+\frac{1}{x}=t\Rightarrow (1-\frac{1}{x^2})dx=dt$
$\therefore I=\int \frac{dt}{t\sqrt{t^2-2}}=\int \frac{dt}{t^2\sqrt{1-\frac{2}{t^2}}}$
Substitute $\frac{\sqrt{2}}{t}=4$ and $\frac{\sqrt{2}}{t^2}dt=du$, we have
$I=\frac{-1}{\sqrt{2}}\int \frac{du}{1-u^2}=\frac{1}{\sqrt{2}}{\cos }^{-1}u+C$ $=\frac{1}{\sqrt{2}}{\cos }^{-1}\left(\frac{\sqrt{2}}{x+\frac{1}{x}}\right)+C$
$=\frac{1}{\sqrt{2}}{\cos }^{-1}\left(\frac{\sqrt{2}x}{x^2+1}\right)+C$ $\therefore k=\frac{1}{\sqrt{2}}$