check convergence $\int _{-\infty }^0\:\frac{e^{3x}}{1+x^2} $ 
check convergence of  $\int _{-\infty }^0\:\frac{e^{3x}}{1+x^2} $

so for $x\le0$ : $e^{3x}\le1$ then we get $0\le\frac{e^{3x}}{1+x^2}\le\frac{1}{1+x^2}\le\frac{1}{x^2}$
and because $\int _{-\infty }^{-1}\:\frac{1}{x^2}$ is convergence we get $\int _{-\infty }^0\:\frac{e^{3x}}{1+x^2} $ is convergence 
is that answer correct 
thanks alot 

Well, you are correct, but your arguments could be a lot more precise.
Note that when: $x \in (-\infty,0), e^x < 1$. Hence, $$f(x) = \frac{e^x}{1+x^2} < \frac{1}{1+x^2} = g(x)$$
Hence, by the comparison test as: $$\int_{-\infty}^{0} \frac{1}{1+x^2}\, dx = \arctan x \bigg \lvert_{-\infty}^{0} = \frac{\pi}{2}$$ which converges. Hence, the original integral also converges.
