Question

$\int \frac{x^2+1}{x^4+1}dx$ is equal to.

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

Answer 1

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

Let $I=\int \frac{x^2+1}{x^4+1}dx$
$=\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}-2+2}dx=\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+2}$
$=\int \frac{dt}{t^2+(\sqrt{2}{)}^2}$, where $t=x-\frac{1}{x}$
$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{t}{\sqrt{2}}\right)+c=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)+c$