Let $\mathrm{f}(x)=\int \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x, x \geq 0$, then $\mathrm{f}(3)-\mathrm{f}(1)$ is equal to

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}(x)=\int \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x, x \geq 0 \ & \mathrm{f}(3)-\mathrm{f}(1)=\int_1^3 \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x=\mathrm{I}(	ext { say }) \end{aligned}$ Put $\sqrt{x}=	an 	heta \Rightarrow \mathrm{d} x=2 	an 	heta \sec ^2 	heta \mathrm{d} 	heta$ When $x=1, 	heta=\frac{\pi}{4}$ and when $x=3, 	heta=\frac{\pi}{3}$ $\begin{aligned} 	herefore \quad I & =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{2 	an ^2 	heta \sec ^2 	heta}{\left(1+	an ^2 	hetaight)^2} d 	heta \ & =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{2 	an ^2 	heta}{1+	an ^2 	heta} d 	heta \ & =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{2 \sin ^2 	heta+\cos ^2 	heta}{} d 	heta \ & =\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}(1-\cos 2 	heta) \mathrm{d} 	heta \ & =\left[	heta-\frac{\sin 2 	heta}{2}ight]_{\frac{\pi}{4}}^{\frac{\pi}{3}} \ & =\left(\frac{\pi}{3}-\frac{\pi}{4}ight)-\frac{1}{2}\left(\frac{\sqrt{3}}{2}-1ight) \ & =\frac{\pi}{12}-\frac{\sqrt{3}}{4}+\frac{1}{2} \end{aligned}$

Let $\mathrm{f}(x)=\int \frac{\sqrt{x}}{(1+x)^2} \mathrm{~d} x, x \geq 0$, then…

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