Question

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

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

Answer 1

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

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