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

User Ute
by
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1 Answer

0 votes

Answer:

Integral doesn't converge

Explanation:

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


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

For the integrand
(x)/(x^2+3) substitute:


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


\lim_k \to -\infty} ((1)/(2)  \int\limits^0_k {(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) ((1)/(2) log(x^2+3) \left \{ {{0} \atop {k}} \right )

Evaluating the limits:


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

Hence, the integral doesn't converge

User James Coote
by
8.4k points
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