Let f(x) = int frac{sqrt{x}}{(1+x)^2} , dx, x | vedclass

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(C) We are given $f(x) = \int \frac{\sqrt{x}}{(1+x)^2} \, dx$. We need to evaluate $f(3) - f(1) = \int_1^3 \frac{\sqrt{x}}{(1+x)^2} \, dx$. Let $\sqrt

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$\frac{\pi}{12} - \frac{\sqrt{3}}{4} + \frac{1}{2}$

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(C) We are given $f(x) = \int \frac{\sqrt{x}}{(1+x)^2} \, dx$. We need to evaluate $f(3) - f(1) = \int_1^3 \frac{\sqrt{x}}{(1+x)^2} \, dx$. Let $\sqrt{x} = 	an 	heta$,then $x = 	an^2 	heta$ and $dx = 2 	an 	heta \sec^2 	heta \, d	heta$. When $x = 1$,$	an 	heta = 1 \Rightarrow 	heta = \frac{\pi}{4}$. When $x = 3$,$	an 	heta = \sqrt{3} \Rightarrow 	heta = \frac{\pi}{3}$. Substituting these into the integral: $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{	an 	heta \cdot 2 	an 	heta \sec^2 	heta}{(1 + 	an^2 	heta)^2} \, d	heta = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{2 	an^2 	heta \sec^2 	heta}{\sec^4 	heta} \, d	heta = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} 2 \sin^2 	heta \, d	heta$. Using the identity $2 \sin^2 	heta = 1 - \cos 2	heta$: $I = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (1 - \cos 2	heta) \, d	heta = [	heta - \frac{\sin 2	heta}{2}]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$. $I = (\frac{\pi}{3} - \frac{\sin(2\pi/3)}{2}) - (\frac{\pi}{4} - \frac{\sin(\pi/2)}{2}) = (\frac{\pi}{3} - \frac{\sqrt{3}}{4}) - (\frac{\pi}{4} - \frac{1}{2})$. $I = \frac{\pi}{12} - \frac{\sqrt{3}}{4} + \frac{1}{2}$.

Let $f(x) = \int \frac{\sqrt{x}}{(1+x)^2} \, dx, x \geq 0$. Then $f(3) - f(1)$ is equal to

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