Answer:
Integral doesn't converge
Step-by-step explanation:
In order to computed the improper integral replace infinite limit with a finite value:
[tex]\lim_k \to -\infty} (\int\limits^0_k {\frac{x}{x^2+3} } \, dx )[/tex]
For the integrand [tex]\frac{x}{x^2+3}[/tex] substitute:
[tex]u=x^2+3\\du=2xdx[/tex]
[tex]\lim_k \to -\infty} (\frac{1}{2} \int\limits^0_k {\frac{1}{u} } \, du )[/tex]
The integral of 1/u is log(u):
Applying the fundamental theorem of calculus and substituing back for u=x^2 +3:
[tex]\lim_{k \to -\infty} (\frac{1}{2} log(x^2+3) \left \{ {{0} \atop {k}} \right )[/tex]
Evaluating the limits:
[tex]\lim_{k \to -\infty} (\frac{1}{2} log(x^2+3) \left \{ {{0} \atop {k}} \right ) =0.549-\infty= \infty[/tex]
Hence, the integral doesn't converge
