Evaluating an Improper Integral :-, Determine whether the improper integral diverges or converges. Evaluate the integral if it converges :- ∫−∞∞416+x2dx

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Evaluating an Improper Integral :-, Determine whether the improper integral diverges or converges. Evaluate the integral if it converges :-  ∫−∞∞416+x2dx

lovagwb · Accepted Answer

Step 1  Given,  ∫−∞∞416+x2dx  Determine whether the improper integral diverges or converges.  Step 2  If either or any one limit point of the integral is infinite then we say that the integral is improper integral.  Consider,  ∫−∞∞416+x2dx  Here we have both the limit points are infinity,  So we say that the integral is improper,  Now Use the Fundamental law of integration we get,  ∫abf(x)dx=F(b)−F(a)  ∫−∞∞416+x2dx  We know that integration formula,  ∫dxa2+x2=1atan−1(xa)+C  ∫442+x2dx=4[14tan−1(x4)]+C  ∫442+x2dx=[tan−1(x4)]+C  Now Use the Fundamental law we write that as,  ∫−∞∞416+x2dx=tan−1(∞4)−tan−1(−∞4)  =tan−1(∞)−tan−1(−∞)  =π2−(−π2)  =π2+π2  =2π2  =π  Therefore The integral is Improper and Converges.

William Appel · Accepted Answer

∫4x2+16dx  Substitution u=x4⇒dudx=14⇒dx=4du:  =∫1616u2+16du  Simplifying:  =∫1u2+1du  This is the well-known tabular integral:  =arctan⁡(u)  Reverse replacement u=x4:  =arctan⁡(x4)  Problem solved:  ∫4x2+16dx  =arctan⁡(x4)+C

nick1337 · Accepted Answer

∫416+x2∗dx We represent the initial integral in the form: ∫416+x2∗dx=∫4x2+16∗dx ∫4x2+16∗dx This is a tabular integral: ∫4x2+16∗dx=arctan⁡(x4)+C

Evaluating an Improper Integral :-, Determine whether the improper integral

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