Answer
$$\left( {\overline x ,\overline y } \right) = \left( {0,\frac{{\pi + 2}}{{4\pi }}} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{1 + {x^2}}},{\text{ }}y = 0,{\text{ }}x = - 1,{\text{ }}x = 1,{\text{ }}\rho \left( {x,y} \right) = k \cr
& {\text{The region }}R{\text{ is}} \cr
& R = \left\{ {\left. {\left( {x,y} \right)} \right|0 \leqslant y \leqslant \frac{1}{{1 + {x^2}}},{\text{ }} - 1 \leqslant x \leqslant 1} \right\} \cr
& {\text{The mass }}m{\text{ of the lamina is given by}} \cr
& m = \iint\limits_R {\rho \left( {x,y} \right)dA} \cr
& m = \int_{ - 1}^1 {\int_0^{\frac{1}{{1 + {x^2}}}} k } dydx \cr
& m = k\int_{ - 1}^1 {\left[ {\int_0^{\frac{1}{{1 + {x^2}}}} {dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& m = k\int_{ - 1}^1 {\left[ y \right]_0^{\frac{1}{{1 + {x^2}}}}} dx \cr
& m = k\int_{ - 1}^1 {\frac{1}{{1 + {x^2}}}} dx \cr
& {\text{Integrate}} \cr
& m = k\left[ {\arctan x} \right]_{ - 1}^1 \cr
& m = k\left[ {\arctan \left( 1 \right) - \arctan \left( { - 1} \right)} \right] \cr
& m = k\left( {\frac{\pi }{4} - \left( { - \frac{\pi }{4}} \right)} \right) \cr
& m = \frac{{k\pi }}{2} \cr
& \cr
& {\text{*The moment of mass with respect to the }}x{\text{ axis is:}} \cr
& {M_x} = \iint\limits_R {y\rho \left( {x,y} \right)dA},{\text{ then}} \cr
& {M_x} = \int_{ - 1}^1 {\int_0^{\frac{1}{{1 + {x^2}}}} {y\left( k \right)} } dydx \cr
& {M_x} = k\int_{ - 1}^1 {\left[ {\int_0^{\frac{1}{{1 + {x^2}}}} {ydy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_x} = k\int_{ - 1}^1 {\left[ {\frac{1}{2}{y^2}} \right]_0^{\frac{1}{{1 + {x^2}}}}} dx \cr
& {M_x} = \frac{k}{2}\int_{ - 1}^1 {\left[ {{{\left( {\frac{1}{{1 + {x^2}}}} \right)}^2}} \right]} dx \cr
& {M_x} = \frac{k}{2}\int_{ - 1}^1 {\frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}}} dx \cr
& {\text{Integrate by a CAS}} \cr
& {M_x} = \frac{k}{4}\left[ {\frac{x}{{{x^2} + 1}} + {{\tan }^{ - 1}}\left( x \right)} \right]_{ - 1}^1 \cr
& {M_x} = \frac{k}{4}\left[ {\frac{1}{2} + {{\tan }^{ - 1}}\left( 1 \right)} \right] - \frac{k}{4}\left[ { - \frac{1}{2} + {{\tan }^{ - 1}}\left( { - 1} \right)} \right] \cr
& {M_x} = \frac{k}{8} + \frac{k}{4}{\tan ^{ - 1}}\left( 1 \right) + \frac{k}{8} + \frac{k}{4}{\tan ^{ - 1}}\left( 1 \right) \cr
& {M_x} = \frac{k}{4} + \frac{k}{2}\left( {\frac{\pi }{4}} \right) \cr
& {M_x} = \frac{k}{4}\left( {1 + \frac{\pi }{2}} \right) \cr
& {M_x} = \frac{{\pi + 2}}{8}k \cr
& \cr
& {\text{*The moment of mass with respect to the }}y{\text{ axis is:}} \cr
& {M_y} = \iint\limits_R {x\rho \left( {x,y} \right)dA},{\text{ then}} \cr
& {M_y} = \int_{ - 1}^1 {\int_0^{\frac{1}{{1 + {x^2}}}} {x\left( k \right)} } dydx \cr
& {M_y} = k\int_{ - 1}^1 {x\left[ {\int_0^{\frac{1}{{1 + {x^2}}}} {dy} } \right]} dx \cr
& {\text{Integrating with respect to }}y \cr
& {M_y} = k\int_{ - 1}^1 {x\left( {\frac{1}{{1 + {x^2}}}} \right)} dx \cr
& {M_y} = k\int_{ - 1}^1 {\frac{x}{{1 + {x^2}}}} dx \cr
& {\text{Integrate}} \cr
& {M_y} = \frac{k}{2}\left[ {\ln \left( {1 + {x^2}} \right)} \right]_{ - 1}^1 \cr
& {M_y} = \frac{k}{2}\left[ {\ln \left( 2 \right) - \ln \left( 2 \right)} \right] \cr
& {M_y} = 0 \cr
& \cr
& {\text{The center of mass is}} \cr
& \left( {\overline x ,\overline y } \right) = \left( {\frac{{{M_y}}}{m},\frac{{{M_x}}}{m}} \right) \cr
& \frac{{{M_y}}}{m} = \frac{0}{{\frac{{k\pi }}{2}}} = 0 \cr
& \frac{{{M_x}}}{m} = \frac{{\frac{{\pi + 2}}{8}k}}{{\frac{{k\pi }}{2}}} = \frac{{\pi + 2}}{8}\left( {\frac{2}{\pi }} \right) = \frac{{\pi + 2}}{{4\pi }} \cr
& \left( {\overline x ,\overline y } \right) = \left( {0,\frac{{\pi + 2}}{{4\pi }}} \right) \cr} $$
