Integration

**guess** · March 8th 2006, 02:50 AM

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

integrate 1/(1+X^4)

Please Help

**Jameson** · March 8th 2006, 04:22 AM

This is a nasty one. I'd suggest starting with trig substitution. Are you familiar with this method. I'm posting Mathematica's solution.

**CaptainBlack** · March 8th 2006, 04:35 AM

Originally Posted by guess

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

integrate 1/(1+X^4)

Please Help

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

RonL

**guess** · March 8th 2006, 05:04 AM

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

**guess** · March 8th 2006, 05:06 AM

so sorry about the duplicate post... hope you don't mind...

**guess** · March 8th 2006, 05:17 AM

i think the solution do not work with upper limit 1 lower limit 0.. how you think so??

**Jameson** · March 8th 2006, 05:28 AM

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

**guess** · March 8th 2006, 05:40 AM

Thanks

I will try to solve but for the solution you provide can't work with upper limit 1 and lower limit 0

**CaptainBlack** · March 8th 2006, 06:38 AM

Originally Posted by guess

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

integrate 1/(1+X^4)

Please Help

Observe that

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

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

**TD!** · March 8th 2006, 06:42 AM

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:

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

**ThePerfectHacker** · March 8th 2006, 02:11 PM

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.

**guess** · March 9th 2006, 06:07 AM

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

**TD!** · March 9th 2006, 07:31 AM

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.

**guess** · March 10th 2006, 03:35 PM

How do i integrate from there??

**TD!** · March 11th 2006, 03:55 AM

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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