How to check if this improper integral converges or diverges ? $$ I=\displaystyle \int_{0}^{\infty} \dfrac{\mathrm{d}x}{x^{2}+\sin x} $$ I took $$f(x)=\dfrac{1}{x}$$ and $$g(x)=\dfrac{1}{x+ \dfrac{\sin x}{x}}$$ then tried to find out limit of f(x)/g(x) as x tends to infinity with this i got the the integral as convergent
but when i went with $$ \dfrac{\sin x}{x} \le 1$$ and got $$ \dfrac{1}{x\left(x+ \dfrac{\sin x}{x}\right)} \ge \dfrac{1}{x(x+1)} \implies \displaystyle \int_{0}^{\infty}\dfrac{1}{x(x+1)} dx =\infty $$
two different results using different method ? please suggest something , thanks 

You did the second example correctly, and you did the first example almost correctly as well, but messed it up at the end.

Theorem (Limit Comparison Test): Suppose thatthere are two functions,
  $f(x)$ and $g(x)$ such that $\lim_{x\to\infty}f(x)/g(x)=c>0$. Then
  $\int_a^\infty f(x)dx$ converges if and only if $\int_a^\infty g(x)dx$
  does.

You correctly computed the limit and found that it is constant. That means that either both functions have convergent integrals or both have divergent integrals. $\int_0^\infty dx/x$ is a divergent integral though, so the correct conclusion to reach with method $1$ is that the integral diverges, not converges.
