Question

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

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

Answer 1

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

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

$\int \frac{1+x^{-2}}{x^2+x^{-2}}.dx$ [divide with $x^2$]

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

$\int \frac{1+x^{-2}}{(x-x^{-1}{)}^2+2}dx$

$u=x-x^{-1}$

$\int \frac{dU}{U^2+2}=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{U}{\sqrt{2}}+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{(x-x^{-1})}{\sqrt{2}}+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)+C$