# Commentary for "The Antiderivative of 1/(1+x^4)."

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Moderator edit: This commentary topic pertains to the following tutorial:

http://mathhelpboards.com/math-notes-49/antiderivative-1-1-x^4-4104.html

Excuse me to introduce another approach

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

$$\displaystyle \frac{1}{2} \int \frac{2-x^2+x^2}{1+x^4} = \frac{1}{2} \int \frac{1-x^2}{1+x^4} + \frac{1+x^2}{1+x^4}$$

Let us take the first integral :

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

$$\displaystyle \frac{1}{2} \int \frac{\frac{1}{x^2}-1}{\frac{1}{x^2}+x^2+2-2} = \frac{1}{2} \int \frac{\frac{1}{x^2}-1}{\left(x+\frac{1}{x} \right)^2 -2} =- \frac{1}{2} \int \frac{du}{u^2 -2}$$

Which can be solved by partial fraction to :

$$\displaystyle \frac{1}{4 \sqrt{2}} \log \left( \frac{x^2+\sqrt{2}x+ 1}{ x^2-\sqrt{2}x+ 1} \right) +C$$

Similarily we have :

$$\displaystyle \frac{1}{2} \int \frac{1+x^2}{1+x^4} = \frac{1}{2} \int \frac{\frac{1}{x^2}+ 1}{\frac{1}{x^2}+x^2 }$$

$$\displaystyle \frac{1}{2} \int \frac{\frac{1}{x^2}+ 1}{\frac{1}{x^2}+x^2 -2 +2 } = \frac{1}{2} \int \frac{\frac{1}{x^2}+ 1}{\left( x-\frac{1}{x} \right)^2 +2} = \frac{1}{2} \int \frac{du}{u^2 +2}$$

which easily computes to :

$$\displaystyle \frac{1}{2\sqrt{2}} \arctan \left( \frac{x-\frac{1}{x}}{ \sqrt{2}} \right) +C$$

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