Calculus II - Improper Integral Problem

BaxterCorner
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Homework Statement



Evaluate the integral: ∫(0 to ∞) [dv/((1+v^2)(1+tan^-1(v))]

Homework Equations



U-substitution, taking limit to evaluate improper integrals

The Attempt at a Solution



http://imgur.com/CjkRF
As you can see in the image, I try u-substitution and then take the integral. I end up with ln(0), though, because arctan(0) = 0. The correct answer is ln(1 + ∏/2), but I'm not sure how to get there.
 
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BaxterCorner said:

Homework Statement



Evaluate the integral: ∫(0 to ∞) [dv/((1+v^2)(1+tan^-1(v))]

Homework Equations



U-substitution, taking limit to evaluate improper integrals

The Attempt at a Solution



http://imgur.com/CjkRF
As you can see in the image, I try u-substitution and then take the integral. I end up with ln(0), though, because arctan(0) = 0. The correct answer is ln(1 + ∏/2), but I'm not sure how to get there.

With your substitution the denominator is ##1+u##, not just ##u##. It works either way, but I would suggest the substitution ##u=1+\arctan v## in the first place.
 
Sorry, I intended to write u = 1 + arctan(v), not u = arctan(v). The 1 goes to zero either way though, so I still have the same problem of getting ln(0).
 
Nevermind, I see what you're saying, that would change the bounds. Thanks!
 
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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