I was investigating convergence of such an integral:
$\int_{1}^\infty $$\frac{dx}{x(1+x)} $
I used comparison test:
$\int_{1}^\infty $$\frac{dx}{x(1+x)} $ < $\int_{1}^\infty $$\frac{dx}{x^2} $
to prove it converges. But for an integral
$\int_{0}^\infty $$\frac{dx}{x(1+x)} $
it does not work as $\int_{0}^\infty $$\frac{dx}{x^2} $ diverges too. Which test should I use then?
Notice that the integrand itself diverges as $x\rightarrow 0$. This gives a clue that the integral might not converge in the given range.
The integral works out to be $\log({\frac{x}{1+x}})$ (check by taking the derivative). Substituting the endpoint $x=0$ gives $\infty$. However, the endpoint $x=1$ leads to a finite value.
If you must use a comparison test, you should find something which diverges and lower bounds the given integral.