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

Question

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.

Answer 1

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

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