71I could do it a rather lengthy way. I summed it down to 01∫ dx(3-x2)2 , which can be evaluated by partial frac.
\hspace{-16}\mathbf{\int_{1}^{2}\frac{1}{\left(\sqrt{2x-x^2}+2\right)^2}dx}
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71
1708Nice Effort Vivek.
\hspace{-16}\mathbf{\int_{1}^{2}\frac{1}{\left(\sqrt{2x-x^2}+2\right)^2}dx}$\\\\\\ $\mathbf{\int_{1}^{2}\frac{1}{\left(\sqrt{1-(x-1)^2}+2\right)^2}dx}$\\\\\\ Now Let $\mathbf{(x-1)=\frac{2t}{1+t^2}\Leftrightarrow dx=\frac{2.(1-t^2)}{(1+t^2)^2}}$\\\\\\ $\mathbf{2.\int_{0}^{1}\frac{1-t^2}{(3+t^2)^2}dt}$\\\\\\ Now Put $\mathbf{t=\sqrt{3}.\tan u \Leftrightarrow dt=\sqrt{3}.\sec^2 u.du}$\\\\\\ $\mathbf{\frac{2\sqrt{3}}{9}\int_{0}^{\frac{\pi}{6}}\left(\cos^2 u-3\sin^2 u\right)du}$\\\\\\ $\mathbf{\frac{2\sqrt{3}}{9}\int_{0}^{\frac{\pi}{6}}\left(2\cos (2u)-1\right)du=\frac{1}{3}-\frac{2\sqrt{3}}{9}\times \frac{\pi}{6}}$\\\\\\ So $\boxed{\boxed{\mathbf{\int_{1}^{2}\frac{1}{\left(\sqrt{2x-x^2}+2\right)^2}dx=\frac{1}{3}-\frac{\pi}{9\sqrt{3}}}}}$
