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

Solution

1. Express the integrand as $\frac{1}{1+x^4}=\frac{1}{2}\left(\frac{x^2+1}{x^4+1}-\frac{x^2-1}{x^4+1}\right)$.
2. Divide the numerator and denominator of each fraction by $x^2$ to obtain $\frac{1}{2}\left(\frac{1+1/x^2}{x^2+1/x^2}-\frac{1-1/x^2}{x^2+1/x^2}\right)$.
3. Substitute $u=x-1/x$ for the first term and $v=x+1/x$ for the second term to evaluate the integrals.
4. Integrate the resulting expressions to arrive at $\frac{1}{4\sqrt{2}}\ln \left|\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right|+\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right)+C$.
5. Hence the answer is (B).