People out there... i need some help on this integration problem...
integrate 1/(1+X^4)
Please Help
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
Observe thatOriginally 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
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:
$\displaystyle \to \frac{{Ax + B}}{{x^2 + \sqrt 2 x + 1}} + \frac{{Cx + D}}{{x^2 - \sqrt 2 x + 1}}$
Have you tried my suggestion? It should give: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.
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.