Improper Integral

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• September 16th 2008, 12:06 PM
andrewsx
Improper Integral
integral (from 0 to infinity) of dx/rootx(x+1)
• September 16th 2008, 12:32 PM
Opalg
When x is large, the integrand is approximately 1/x, so the integral will diverge.

If that explanation isn't rigorous enought for you, you can integrate $\int\frac{dx}{\sqrt{x(x+1)}} = \int\frac{2dx}{\sqrt{(2x+1)^2-1}}$ by making the substitution $2x+1=\sec\theta$.