# Determine if the integral converges: $\int_{-\infty}^{2} \frac{e^{3x}dx}{1+x^2}$

Determine if the integral converges:

$$\int_{-\infty}^{2} \frac{e^{3x}dx}{1+x^2}$$

I've tried this:

$$f(x)=\frac{e^{3x}}{1+x^2}>\frac{e^{\ln{3x}}}{1+x^2}=\frac{3x}{1+x^2}=g(x)$$

now since g(x) is similar to $\frac1x$ the integral diverges. [Sparing the limits prof].

Anything i did wrong? Feels to me like i didn't need to pay attetion to $-\infty$, and i'm probably missing something.

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Your problem is that the integral of $g$ goes to minus infinity and not plus infinity. –  N.U. Apr 11 '13 at 18:19

The integral converges. To see this, do a substitution $x=-y$, and then compare with

$$\int_2^{\infty} dy \: e^{-3 y}$$

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felt like i'm missing exactly that, Didn't think about substitution. Thanks ! –  StationaryTraveller Apr 11 '13 at 18:08