# Integration

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• Mar 8th 2006, 03:50 AM
guess
Integration
People out there... i need some help on this integration problem...

integrate 1/(1+X^4)

• Mar 8th 2006, 05:22 AM
Jameson
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, 05:35 AM
CaptainBlack
Quote:

Originally Posted by guess
People out there... i need some help on this integration problem...

integrate 1/(1+X^4)

Duplicate post deleted, because I'm not in as good a mood
as Topsquark this morning - even though it's afternoon here.

RonL
• Mar 8th 2006, 06:04 AM
guess

Very much appreciated with your help
• Mar 8th 2006, 06:06 AM
guess
so sorry about the duplicate post... hope you don't mind...
• Mar 8th 2006, 06:17 AM
guess
i think the solution do not work with upper limit 1 lower limit 0.. how you think so??
• Mar 8th 2006, 06:28 AM
Jameson
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 $\sqrt{1+x^4}$

http://tutorial.math.lamar.edu/AllBr...stitutions.asp
• Mar 8th 2006, 06:40 AM
guess
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, 07:38 AM
CaptainBlack
Quote:

Originally Posted by guess
People out there... i need some help on this integration problem...

integrate 1/(1+X^4)

Observe that

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

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

Now resolve into partial fractions and integrate.

RonL
• Mar 8th 2006, 07:42 AM
TD!
Without working complex, use the radical factorization of 1+x^4:

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

Then do a partial fraction decomposition:

$\to \frac{{Ax + B}}{{x^2 + \sqrt 2 x + 1}} + \frac{{Cx + D}}{{x^2 - \sqrt 2 x + 1}}$
• Mar 8th 2006, 03:11 PM
ThePerfectHacker
Quote:

Originally Posted by TD!
Without working complex, use the radical factorization of 1+x^4:

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

Then do a partial fraction decomposition:

$\to \frac{{Ax + B}}{{x^2 + \sqrt 2 x + 1}} + \frac{{Cx + D}}{{x^2 - \sqrt 2 x + 1}}$

Nice move TD!
You really are the master of analysis.
• Mar 9th 2006, 07:07 AM
guess
can someone please provide a step by step solution to the question?? can't really solve it
• Mar 9th 2006, 08:31 AM
TD!
Quote:

Originally Posted by guess
can someone please provide a step by step solution to the question?? can't really solve it

Have you tried my suggestion? It should give:

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

Now you can integrate both terms, each term will give an ln-part and an arctan-part.
• Mar 10th 2006, 04:35 PM
guess
How do i integrate from there??
• Mar 11th 2006, 04:55 AM
TD!
Let me name both integrals

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

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.

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

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

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

So we have

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

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.

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

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

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

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.

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

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

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

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

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

Combining these

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

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

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

So we have found

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

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.
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