Question

Evaluate $\int \frac{x^2+1}{x^4+1}dx.$

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

Answer 1

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

$I=\int \frac{x^2+1}{x^4+1}dx\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx=\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+2}dx$
Let $x-\frac{1}{x}=t\text{or}d(x-\frac{1}{x})=dt\text{or}(1+\frac{1}{x^2})dx=dt$
$\therefore I=\int \frac{dt}{t^2+{\left(\sqrt{2}\right)}^2}=\frac{1}{\sqrt{2}}{\tan }^{-1}t+C$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C$ $=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)+C$