Question

$\int \frac{x^2-1}{x^4+1}dx$ is equal to (where $C$ is integration constant)

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

Answer 1

The correct option is C $\frac{1}{2\sqrt{2}}\ln \left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+C$

$I=\int \frac{x^2-1}{x^4+1}dx=\int \frac{1-\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx=\int \frac{1-\frac{1}{x^2}}{{(x+\frac{1}{x})}^2-2}dx$
Let, $x+\frac{1}{x}=t$
$\Rightarrow (1-\frac{1}{x^2})dx=dt\therefore I=\int \frac{dt}{t^2-{\left(\sqrt{2}\right)}^2}=\frac{1}{2\sqrt{2}}\ln \left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|+C\{\because \int \frac{1}{t^2-a^2}dt=\frac{1}{2a}\ln \left|\frac{t-a}{t+a}\right|+C\}=\frac{1}{2\sqrt{2}}\ln \left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right|+C$