int_0^infty frac{x^3 , dx}{(x^2 + 4)^2} = | vedclass

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Question

(B) Let $I = \int_0^\infty \frac{x^3 \, dx}{(x^2 + 4)^2}$.Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$.When $x = 0

Vedclass · Accepted Answer

$\infty$

Vedclass · Answer

(B) Let $I = \int_0^\infty \frac{x^3 \, dx}{(x^2 + 4)^2}$.Substitute $t = x^2$,then $dt = 2x \, dx$,which implies $x \, dx = \frac{dt}{2}$.When $x = 0$,$t = 0$. When $x 	o \infty$,$t 	o \infty$.$I = \int_0^\infty \frac{t \cdot (x \, dx)}{(t + 4)^2} = \int_0^\infty \frac{t \cdot \frac{dt}{2}}{(t + 4)^2} = \frac{1}{2} \int_0^\infty \frac{t}{(t + 4)^2} \, dt$.We can rewrite the integrand as $\frac{t}{(t + 4)^2} = \frac{t + 4 - 4}{(t + 4)^2} = \frac{1}{t + 4} - \frac{4}{(t + 4)^2}$.$I = \frac{1}{2} \left[ \int_0^\infty \frac{1}{t + 4} \, dt - \int_0^\infty \frac{4}{(t + 4)^2} \, dt ight]$.$I = \frac{1}{2} \left[ \ln(t + 4) + \frac{4}{t + 4} ight]_0^\infty$.As $t 	o \infty$,$\ln(t + 4) 	o \infty$ and $\frac{4}{t + 4} 	o 0$.Thus,the integral diverges to $\infty$.

$\int_0^\infty \frac{x^3 \, dx}{(x^2 + 4)^2} = $

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