Integral of 1/(x^4+1) dx: different ways of doing this?
- Thread starter Zenith
- Start date
Re: Integral
Yes, there are many ways. There is a method from a while back that Soroban posted. Here: viewtopic.php?f=3&t=28403&p=108823&hilit=+weird#p108823
Here is mine. But as I remember, it is not a nice as the one he posted. This is just partial fractions. Makes it rather arduous.
\(\displaystyle \int\frac{1}{1+x^{4}}dx\)
\(\displaystyle \frac{1}{4}\int\frac{2-\sqrt{2}x}{x^{2}-\sqrt{2}x+1}+\frac{1}{4}\int\frac{2+\sqrt{2}x}{x^{2}+\sqrt{2}x+1}dx\)
\(\displaystyle \frac{-\sqrt{2}}{8}\int\frac{2x-\sqrt{2}}{x^{2}-\sqrt{2}x+1}+\frac{1}{4}\int\frac{1}{x^{2}-\sqrt{2}x+1}dx+\frac{1}{4}\int\frac{2+\sqrt{2}x}{x^{2}+\sqrt{2}x+1}dx\)
Now, for the first one, let \(\displaystyle u=x^{2}-\sqrt{2}x+1, \;\ du=(2x-\sqrt{2})dx\)
Then we get \(\displaystyle \frac{-\sqrt{2}}{8}\int\frac{1}{u}du\)
Which gives \(\displaystyle \frac{-\sqrt{2}}{8}ln(x^{2}-\sqrt{2}x+1)\)
Perhaps you can plod through the rest. They will involve arctan and ln
As I said, this is not the cleverest method, but it works.
Yes, there are many ways. There is a method from a while back that Soroban posted. Here: viewtopic.php?f=3&t=28403&p=108823&hilit=+weird#p108823
Here is mine. But as I remember, it is not a nice as the one he posted. This is just partial fractions. Makes it rather arduous.
\(\displaystyle \int\frac{1}{1+x^{4}}dx\)
\(\displaystyle \frac{1}{4}\int\frac{2-\sqrt{2}x}{x^{2}-\sqrt{2}x+1}+\frac{1}{4}\int\frac{2+\sqrt{2}x}{x^{2}+\sqrt{2}x+1}dx\)
\(\displaystyle \frac{-\sqrt{2}}{8}\int\frac{2x-\sqrt{2}}{x^{2}-\sqrt{2}x+1}+\frac{1}{4}\int\frac{1}{x^{2}-\sqrt{2}x+1}dx+\frac{1}{4}\int\frac{2+\sqrt{2}x}{x^{2}+\sqrt{2}x+1}dx\)
Now, for the first one, let \(\displaystyle u=x^{2}-\sqrt{2}x+1, \;\ du=(2x-\sqrt{2})dx\)
Then we get \(\displaystyle \frac{-\sqrt{2}}{8}\int\frac{1}{u}du\)
Which gives \(\displaystyle \frac{-\sqrt{2}}{8}ln(x^{2}-\sqrt{2}x+1)\)
Perhaps you can plod through the rest. They will involve arctan and ln
As I said, this is not the cleverest method, but it works.
