What is the Integral of 1/(x^4+1) from 0 to b?

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Homework Statement


i do not know where to start with this one can anyone point me where to go please?

\int \frac{1}{x^{4}+1}

lower limit = 0 upper = infinity


Homework Equations



like i said i don't know where to go with this one.
i mean i know I am going to have to do integrate from 0 to b but i don't know where to start here.

The Attempt at a Solution

 
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It seems really messy if you integrate using substitutions ... etc.

I would recommend using f = 1/(x^2.y^2+1)

some MATLAB mess

>> f = 1/(x^4+1)

f =

1/(x^4+1)


>> int(f,x,0,inf)

ans =

1/4*pi*2^(1/2)


>> int(f,x)

ans =

1/8*2^(1/2)*log((x^2+x*2^(1/2)+1)/(x^2-x*2^(1/2)+1))+1/4*2^(1/2)*atan(x*2^(1/2)+1)+1/4*2^(1/2)*atan(x*2^(1/2)-1)
 
o wait i think i need to compare i think i have it ill repost. thanks rootX seeing your answer made me see that we have yet to really learn that, so it can't be. thanks again. ill post with what i get to see if I am right.
 
Here some hints to get you going. You'll have to decompose it to partial fractions, complete a square, and then use the derivative for arctan f(x). It's quite tedious, but most certainly doable.
 


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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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