The value of the integral int_0^{infty} frac{ | vedclass

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(B) Let $I = \int_0^{\infty} \frac{dx}{(1+x^2)(1+x)^2}$.Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.When $x = 0$,$	heta = 0$,an

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$\frac{1}{2}$

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(B) Let $I = \int_0^{\infty} \frac{dx}{(1+x^2)(1+x)^2}$.Substitute $x = 	an 	heta$,then $dx = \sec^2 	heta \, d	heta$.When $x = 0$,$	heta = 0$,and when $x 	o \infty$,$	heta = \frac{\pi}{2}$.$I = \int_0^{\pi/2} \frac{\sec^2 	heta \, d	heta}{(1+	an^2 	heta)(1+	an 	heta)^2} = \int_0^{\pi/2} \frac{\sec^2 	heta \, d	heta}{\sec^2 	heta (1+	an 	heta)^2} = \int_0^{\pi/2} \frac{d	heta}{(1+	an 	heta)^2}$.Using the property $\int_0^a f(	heta) \, d	heta = \int_0^a f(a-	heta) \, d	heta$,we have:$I = \int_0^{\pi/2} \frac{d	heta}{(1+	an(\frac{\pi}{2}-	heta))^2} = \int_0^{\pi/2} \frac{d	heta}{(1+\cot 	heta)^2} = \int_0^{\pi/2} \frac{	an^2 	heta \, d	heta}{(	an 	heta + 1)^2}$.Adding the two expressions for $I$:$2I = \int_0^{\pi/2} \frac{1 + 	an^2 	heta}{(1+	an 	heta)^2} \, d	heta = \int_0^{\pi/2} \frac{\sec^2 	heta}{(1+	an 	heta)^2} \, d	heta$.Let $u = 1 + 	an 	heta$,then $du = \sec^2 	heta \, d	heta$.$2I = \int_1^{\infty} \frac{du}{u^2} = \left[ -\frac{1}{u} ight]_1^{\infty} = 0 - (-1) = 1$.Therefore,$I = \frac{1}{2}$.

The value of the integral $\int_0^{\infty} \frac{dx}{(1+x^2)(1+x)^2}$ is

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