Need help with comparison method for improper integral?

mottov2
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Homework Statement


\int \frac{arctan(x)}{2+e^x}dx
where the interval of the integrand is from 0 to infinity.

In order to use the comparison method I need to compare 2 functions but I am having so much difficulty figuring out what function to compare it to.

Its not just this particular question, any question asking to use comparison method, I can't seem to figure out the function to compare it to.

Can I get some tips to finding a function to compare it to?
 
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tan-1(x) < 2 for all x, so therefore...

\frac{tan^{-1}(x)}{2 + e^{x}} &lt; \frac{2}{2 + e^{x}}

for all x (Since 2+ex is never 0, one doesn't have to worry about division by zero.)

So try using the comparison test with that.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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