What Substitution Solves This Integral?

sleepwalker27
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1.
http://www.imageurlhost.com/images/cnj1t05jh6e4fxqy4i5_integral.png
I know that this integral is solved by the sustitution method

The Attempt at a Solution


I tried converting the integral to the form of Arctanx, but that x2 on the numerator ruined everything. Thanks
 
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sleepwalker27 said:
1.
http://www.imageurlhost.com/images/cnj1t05jh6e4fxqy4i5_integral.png
I know that this integral is solved by the sustitution method

The Attempt at a Solution


I tried converting the integral to the form of Arctanx, but that x2 on the numerator ruined everything. Thanks
Please show us the substitution you used, which sounds like a trig substitution.
 
Mark44 said:
Please show us the substitution you used, which sounds like a trig substitution.
I tried to made the integral to the form ∫dx/x2+1 so that their solution is something like arctanx
 
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sleepwalker27 said:
I tried to made the integral to the form ∫dx/x2+1 so that their solution is something like arctanx
I figured you did something like that, but that isn't what I asked you for. Please show me your substitution.
 
Did you not notice that the numerator and denominator have the same degree?

\frac{2x^2}{2x^2+ 1}= 1- \frac{1}{2x^2+ 1}
 
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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