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!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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