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 
 A: 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.
A: Your answer is correct, as you did the given integral may be obtained as a sum of two convergent integrals,
$$
\int _{-\infty }^0\:\frac{e^{3x}}{1+x^2} \:dx=\int _{-\infty }^{-1}\:\frac{e^{3x}}{1+x^2}\:dx +\int _{-1 }^0\:\frac{e^{3x}}{1+x^2}\:dx,
$$ the integrand being continuous over $(-\infty,\infty)$.
A: Yes it is correct ! You only have to check the convergence at infinity (as you did) since the fonction is well defined at $0$.
A: $f(x)=\frac{e^{-3x}}{1+x^2}$ is a positive function on $\mathbb{R}^+$ and
$$\int_{0}^{+\infty}\frac{e^{-3x}}{1+x^2}\,dx\stackrel{\text{Cauchy-Schwarz}}{\leq}\sqrt{\int_{0}^{+\infty}e^{-6x}\,dx \int_{0}^{+\infty}\frac{dx}{(1+x^2)^2}}=\sqrt{\frac{\pi}{24}}. $$
This inequality can also be improved up to
$$\int_{0}^{+\infty}\frac{e^{-3x}}{1+x^2}\,dx\stackrel{\text{Cauchy-Schwarz}}{\leq}\sqrt{\int_{0}^{+\infty}(1+x)^3e^{-6x}\,dx \int_{0}^{+\infty}\frac{dx}{(1+x^2)^2 (1+x)^3}}=\sqrt{\frac{61(8-\pi)}{3456}}=\color{green}{0.29}2836\ldots $$
