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Basile [38]
3 years ago
5

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

Mathematics
1 answer:
marshall27 [118]3 years ago
6 0

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

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