Welcome to our community

Be a part of something great, join today!

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


Well-known member
MHB Math Helper
Jan 17, 2013
Moderator edit: This commentary topic pertains to the following tutorial:


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 :

\(\displaystyle \frac{1}{2\sqrt{2}} \arctan \left( \frac{x-\frac{1}{x}}{ \sqrt{2}} \right) +C \)
Last edited by a moderator: