How can trigonometric substitution be used to simplify a complex integral?

hyper
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I should probably mention that the answer is supposed to be:

2*arctan(2x)+4x/(4x^2+1) +C
 
I didn't go through that in detail but it looks like a very strange way to attack the problem! You have a square of a square and you write it as a fourth power of a square root of a square so you can apply a trig substitution!
You don't need the square root to apply a trig substitution. Let 2x= tan t and 4x2+ 1= tan2 t+ 1= sec2. (4x2+ 1)2= sec4 t and 2dx= sec2 t dt. Your integral becomes
\int\frac{8dx}{(4x^2+ 1)^2}= \int \frac{4dt}{sec^2 t}= 4\int cos^2 t dt[/itex]<br /> That should be easy.
 
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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