Tips for Solving a Difficult Integral

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Homework Statement



Hi, I just wondered if anyone could give me any tips on what what I consider a very hard integral:

\int (\sqrt{ 1/2 + x^6 + 1/16x^6})dx

Homework Equations



Integration by substitution? i.e u = some form of x

The Attempt at a Solution



I've been looking at this integral for about 2 days now and have got pretty much nowhere with it. I'm finding it that difficult that I'm beginning to think I've done the work up to this integral incorrectly (despite the fact that I've checked it a hundred times).
The only thing I can think of is that some sort of substitution is needed, but what to substitute for I have no idea.

If any could give me any tips on how I can advance with this integral I would very much appreciate it.

Thanks in advance,

Kam
 
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Is that 1/(16x^6) or (1/16)x^6 ?
 
If you are given the limits of integration then I would use a numerical technique (Simpson's Rule would work). It's impractical to symbolically work out crazy integrals.
 
gneill said:
Is that 1/(16x^6) or (1/16)x^6 ?
If it's the latter, the integral is pretty straightforward. Start by putting everything in the radical over a common denominator.
 
gneill said:
Is that 1/(16x^6) or (1/16)x^6 ?

It is 1/(16x^6)
 
mrmiller1 said:
If you are given the limits of integration then I would use a numerical technique (Simpson's Rule would work). It's impractical to symbolically work out crazy integrals.

Yes the limits are 2 and 1, I have never heard of Simpson's rule but I will check it out.
 
1/2 + x^6 +1/(16x^6)=\frac {16x^{12}+8x^6+1}{16x^6}=\frac {(4x^6+1)^2}{(4x^3)^2}
 
Delta² said:
1/2 + x^6 +1/(16x^6)=\frac {16x^{12}+8x^6+1}{16x^6}=\frac {(4x^6+1)^2}{(4x^3)^2}

Okay thank you for that Delta, I will now try and see if I can find a sensible substitution / another method to solve the integral from there.
 
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