People out there... i need some help on this integration problem...

integrate 1/(1+X^4)

Please Help

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- Mar 8th 2006, 02:50 AMguessIntegration
People out there... i need some help on this integration problem...

integrate 1/(1+X^4)

Please Help - Mar 8th 2006, 04:22 AMJameson
This is a nasty one. I'd suggest starting with trig substitution. Are you familiar with this method. I'm posting Mathematica's solution.

- Mar 8th 2006, 04:35 AMCaptainBlackQuote:

Originally Posted by**guess**

as Topsquark this morning - even though it's afternoon here.

RonL - Mar 8th 2006, 05:04 AMguess
Thanks a lot Jameson. I'm not very familiar about this method trigo substitution. can you further explain about it or do you have any website to intro about this method?

Very much appreciated with your help - Mar 8th 2006, 05:06 AMguess
so sorry about the duplicate post... hope you don't mind...

- Mar 8th 2006, 05:17 AMguess
i think the solution do not work with upper limit 1 lower limit 0.. how you think so??

- Mar 8th 2006, 05:28 AMJameson
Here's a website I found with some example of trig substitution. Draw a triangle, with one leg x^2, the other leg 1, and the hypotenuse as $\displaystyle \sqrt{1+x^4}$

http://tutorial.math.lamar.edu/AllBr...stitutions.asp - Mar 8th 2006, 05:40 AMguess
Thanks

I will try to solve but for the solution you provide can't work with upper limit 1 and lower limit 0 - Mar 8th 2006, 06:38 AMCaptainBlackQuote:

Originally Posted by**guess**

$\displaystyle

\frac{1}{1+x^4}=\frac{1}{(x-\omega_1)(x-\omega_2)(x-\omega_3)(x-\omega_4)}

$

where $\displaystyle \omega_1,\ \omega_2,\ \omega_3$ and $\displaystyle \omega_4$ are the four complex roots of $\displaystyle -1$.

Now resolve into partial fractions and integrate.

RonL - Mar 8th 2006, 06:42 AMTD!
Without working complex, use the radical factorization of 1+x^4:

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

Then do a partial fraction decomposition:

$\displaystyle \to \frac{{Ax + B}}{{x^2 + \sqrt 2 x + 1}} + \frac{{Cx + D}}{{x^2 - \sqrt 2 x + 1}}$ - Mar 8th 2006, 02:11 PMThePerfectHackerQuote:

Originally Posted by**TD!**

You really are the master of analysis. - Mar 9th 2006, 06:07 AMguess
can someone please provide a step by step solution to the question?? can't really solve it

- Mar 9th 2006, 07:31 AMTD!Quote:

Originally Posted by**guess**

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

Now you can integrate both terms, each term will give an ln-part and an arctan-part. - Mar 10th 2006, 03:35 PMguess
How do i integrate from there??

- Mar 11th 2006, 03:55 AMTD!
Let me name both integrals

$\displaystyle I = \int {\frac{{\sqrt 2 \left( {\sqrt 2 - x} \right)}}{{4\left( {x^2 - \sqrt 2 x + 1} \right)}}dx} ,J = \int {\frac{{\sqrt 2 \left( {\sqrt 2 + x} \right)}}{{4\left( {x^2 + \sqrt 2 x + 1} \right)}}dx} $

I'll integrate one term, the other one goes completely the same. I'll do I.

As I said, we'll be getting an ln and an inverse tangent. To get the ln part, we find the derivative of the denominator.

$\displaystyle \left( {x^2 - \sqrt 2 x + 1} \right)^\prime = 2x - \sqrt 2 $

We manipulate the nominator to get this derivative and split the integral.

$\displaystyle \int {\frac{{\sqrt 2 \left( {\sqrt 2 - x} \right)}}{{4\left( {x^2 - \sqrt 2 x + 1} \right)}}dx} = - \frac{{\sqrt 2 }}{8}\int {\frac{{2x - \sqrt 2 - \sqrt 2 }}{{x^2 - \sqrt 2 x + 1}}dx}$

So we have

$\displaystyle - \frac{{\sqrt 2 }}{8}\int {\frac{{2x - \sqrt 2 }}{{x^2 - \sqrt 2 x + 1}}dx} + \frac{1}{4}\int {\frac{1}{{x^2 - \sqrt 2 x + 1}}dx} $

Now the first integral has exactly the derivative of the denominator in the nominator, which gives the natural logarithm of the denominator as primitive function.

$\displaystyle - \frac{{\sqrt 2 }}{8}\int {\frac{{2x - \sqrt 2 }}{{x^2 - \sqrt 2 x + 1}}dx} = - \frac{{\sqrt 2 }}{8}\ln \left| {x^2 - \sqrt 2 x + 1} \right| + C$

I'll continue with the second integral, which will give an inverse tangent as I said. We'll complete the square in the denominator.

$\displaystyle \frac{1}{4}\int {\frac{1}{{x^2 - \sqrt 2 x + 1}}dx} = \frac{1}{4}\int {\frac{1}{{\left( {x - \frac{{\sqrt 2 }}{2}} \right)^2 + \frac{1}{2}}}dx}$

But for an inverse tangent, we need something of the form (*)²+1, so a +1 instead of +1/2. We'll fix that by bringing the factor 1/2 for the integral and correcting the square.

$\displaystyle \frac{1}{4}\int {\frac{1}{{\frac{1}{2}\left( {2\left( {x - \frac{{\sqrt 2 }}{2}} \right)^2 + 1} \right)}}dx} = \frac{1}{2}\int {\frac{1}{{\left( {\sqrt 2 x - 1} \right)^2 + 1}}dx} $

We're almost there, we just have to integrate with respect to $\displaystyle {\sqrt 2 x - 1}$ instead of x. I'll adjust the dx and correct before the integral.

$\displaystyle \frac{1}{2}\int {\frac{1}{{\left( {\sqrt 2 x - 1} \right)^2 + 1}}dx} = \frac{1}{{2\sqrt 2 }}\int {\frac{1}{{\left( {\sqrt 2 x - 1} \right)^2 + 1}}d\left( {\sqrt 2 x - 1} \right)} $

Now it's a standard integral with the inverse tangent as primitive function.

$\displaystyle \frac{1}{{2\sqrt 2 }}\int {\frac{1}{{\left( {\sqrt 2 x - 1} \right)^2 + 1}}d\left( {\sqrt 2 x - 1} \right)} = \frac{{\sqrt 2 }}{4}\arctan \left( {\sqrt 2 x - 1} \right)$

Combining these

$\displaystyle I = \frac{{\sqrt 2 }}{4}\arctan \left( {\sqrt 2 x - 1} \right) - \frac{{\sqrt 2 }}{8}\ln \left| {x^2 - \sqrt 2 x + 1} \right| + C$

Completely analogous (only the signs change), we find for J

$\displaystyle J = \frac{{\sqrt 2 }}{4}\arctan \left( {\sqrt 2 x + 1} \right) + \frac{{\sqrt 2 }}{8}\ln \left| {x^2 + \sqrt 2 x + 1} \right| + C$

So we have found

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

I don't usually do such long solutions, but I found this a very good example to illustrate this technique which is often used for these kind of integrals. Most of it can be done slightly differently by using substitutions, but I found this a nice way to illustrate how it can be done by only manipulating the integrals.