Question

Find the value of the integral that converges. ∫^0_-[infinity] x/x^2+3 dx.

Answer 1

Answer:

Integral doesn't converge

Step-by-step explanation:

In order to computed the improper integral replace infinite limit with a finite value:

$\lim_k \to -\infty} (\int\limits^0_k {\frac{x}{x^2+3} } \, dx )$

For the integrand $\frac{x}{x^2+3}$ substitute:

$u=x^2+3\\du=2xdx$

$\lim_k \to -\infty} (\frac{1}{2} \int\limits^0_k {\frac{1}{u} } \, du )$

The integral of 1/u is log(u):

Applying the fundamental theorem of calculus and substituing back for u=x^2 +3:

$\lim_{k \to -\infty} (\frac{1}{2} log(x^2+3) \left \{ {{0} \atop {k}} \right )$

Evaluating the limits:

$\lim_{k \to -\infty} (\frac{1}{2} log(x^2+3) \left \{ {{0} \atop {k}} \right ) =0.549-\infty= \infty$

Hence, the integral doesn't converge