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

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(D) We need to evaluate $f(3) - f(1) = \int_{1}^{3} \frac{\sqrt{x}}{(1+x)^2} dx$. Let $\sqrt{x} = t$,then $x = t^2$ and $dx = 2t dt$. When $x=1$,$t=1$

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

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(D) We need to evaluate $f(3) - f(1) = \int_{1}^{3} \frac{\sqrt{x}}{(1+x)^2} dx$. Let $\sqrt{x} = t$,then $x = t^2$ and $dx = 2t dt$. When $x=1$,$t=1$. When $x=3$,$t=\sqrt{3}$. Substituting these into the integral: $f(3) - f(1) = \int_{1}^{\sqrt{3}} \frac{t \cdot 2t}{(1+t^2)^2} dt = 2 \int_{1}^{\sqrt{3}} \frac{t^2}{(1+t^2)^2} dt$. Using integration by parts,let $u = t$ and $dv = \frac{t}{(1+t^2)^2} dt$. Then $du = dt$ and $v = -\frac{1}{2(1+t^2)}$. $2 \int \frac{t^2}{(1+t^2)^2} dt = 2 \left[ -\frac{t}{2(1+t^2)} + \int \frac{1}{2(1+t^2)} dt ight] = -\frac{t}{1+t^2} + 	an^{-1}(t)$. Evaluating from $1$ to $\sqrt{3}$: $f(3) - f(1) = \left[ -\frac{\sqrt{3}}{1+3} + 	an^{-1}(\sqrt{3}) ight] - \left[ -\frac{1}{1+1} + 	an^{-1}(1) ight]$. $= \left( -\frac{\sqrt{3}}{4} + \frac{\pi}{3} ight) - \left( -\frac{1}{2} + \frac{\pi}{4} ight) = \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{\pi}{12}$.

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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