A problem asked by Vijay Kumar Chavan

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

Answer:

$\frac{1}{1 + x^4} = \frac{x^2 + 1}{2(x^4 + 1)} - \frac{x^2 - 1}{2(x^4 + 1)} $

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

= $ \frac {d(1 - \frac{1}{x})} {2((x - \frac{1}{x})^2 + 2)} - \frac {d(1 + \frac{1}{x})} {2((x + \frac{1}{x})^2 - 2)} $

To evaluate $\int \frac {d(1 - \frac{1}{x})} {2((x - \frac{1}{x})^2 + 2)} $,

we substitute $u= x - \frac{1}{x} $, and we get

$\int \frac{du}{2(u^2 + 2)}$

Now substitute $u = \sqrt{2} \tan \theta $, and we get $du = \sqrt{2} \sec^2 \theta \; d\theta$

$\int \frac{du}{2(u^2 + 2)} = \int \frac{d\theta}{2\sqrt{2}} $

= $ \frac{1}{2\sqrt{2}} \theta = \frac{1}{2\sqrt{2}} \arctan \left( \frac{u}{\sqrt{2}} \right)$

= $\frac{1}{2\sqrt{2}} \arctan \left( \frac{ x - \frac{1}{x} }{\sqrt{2}} \right)$

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

To evaluate $\int \frac {d(1 + \frac{1}{x})} {2((x + \frac{1}{x})^2 - 2)} $

we substitute $u= x + \frac{1}{x} $, and we get

$\int \frac{du}{2(u^2 - 2)} = \frac{1}{4\sqrt{2}} \int \left( \frac{1}{u - \sqrt{2}} - \frac{1}{u + \sqrt{2}}\right) \; du $

= $ \frac{1}{4\sqrt{2}} \ln \frac{u+ \sqrt{2}}{u- \sqrt{2}} $

= $ \frac{1}{4\sqrt{2}} \ln \frac{x + \frac{1}{x}+ \sqrt{2}}{x + \frac{1}{x}- \sqrt{2}} $

= $ \frac{1}{4\sqrt{2}} \ln \frac{x^2 + 1+ \sqrt{2}x}{x^2 + 1- \sqrt{2}x} $

So, $\int \frac{1}{1 + x^4} \; dx = \frac{1}{2\sqrt{2}} \arctan \left( \frac{ x^2 - 1 }{\sqrt{2}x} \right) - \frac{1}{4\sqrt{2}} \ln \frac{x^2 + 1+ \sqrt{2}x}{x^2 + 1- \sqrt{2}x} + C$