Our aim is to apply Green's Theorem. Let us take \(\displaystyle{L}{\left({x},{y}\right)}={\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\) and
\(\displaystyle{M}{\left({x},{y}\right)}={\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\). It follows that
\(\displaystyle{\frac{{\partial{L}}}{{\partial{y}}}}={\frac{{\partial}}{{\partial{y}}}}{\left({\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\right)}\)

\(\displaystyle{\frac{{{\frac{{\partial}}{{\partial{y}}}}{\left({y}\right)}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}-{\frac{{\partial}}{{\partial{y}}}}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}{y}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\)

\(\displaystyle={\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right.}}}}\)

\(\displaystyle{\frac{{\partial{M}}}{{\partial{x}}}}={\frac{{\partial}}{{\partial{x}}}}{\left({\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\right)}\)

\(\displaystyle{\frac{{{\frac{{\partial}}{{\partial{x}}}}{\left({x}-{1}\right)}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}-{\frac{{\partial}}{{\partial{x}}}}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}{\left({x}-{1}\right)}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\)

\(\displaystyle={\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\) Let D be the interior of C, that is, D is a circular disk given by \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}\leq{27}\). By Green's Theorem, the given line integral can be obtained as \(\displaystyle\oint_{{{C}}}{L}{\left.{d}{x}\right.}+{M}{\left.{d}{y}\right.}=\int\int_{{{D}}}{\left({\frac{{\partial{M}}}{{\partial{x}}}}-{\frac{{\partial{L}}}{{\partial{y}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle\int\int_{{{D}}}{\left({\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}-{\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle\int\int_{{{D}}}{0}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={0}\) Therefore, the value of the line integrals is \(\displaystyle\oint_{{{C}}}{\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}{\left.{d}{x}\right.}+{\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}{\left.{d}{y}\right.}={0}\)

\(\displaystyle{\frac{{{\frac{{\partial}}{{\partial{y}}}}{\left({y}\right)}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}-{\frac{{\partial}}{{\partial{y}}}}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}{y}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\)

\(\displaystyle={\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right.}}}}\)

\(\displaystyle{\frac{{\partial{M}}}{{\partial{x}}}}={\frac{{\partial}}{{\partial{x}}}}{\left({\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}\right)}\)

\(\displaystyle{\frac{{{\frac{{\partial}}{{\partial{x}}}}{\left({x}-{1}\right)}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}-{\frac{{\partial}}{{\partial{x}}}}{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}{\left({x}-{1}\right)}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\)

\(\displaystyle={\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\) Let D be the interior of C, that is, D is a circular disk given by \(\displaystyle{x}^{{{2}}}+{y}^{{{2}}}\leq{27}\). By Green's Theorem, the given line integral can be obtained as \(\displaystyle\oint_{{{C}}}{L}{\left.{d}{x}\right.}+{M}{\left.{d}{y}\right.}=\int\int_{{{D}}}{\left({\frac{{\partial{M}}}{{\partial{x}}}}-{\frac{{\partial{L}}}{{\partial{y}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle\int\int_{{{D}}}{\left({\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}-{\frac{{{\left({x}-{1}\right)}^{{{2}}}-{4}{y}^{{{2}}}}}{{{\left({\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}\right)}^{{{2}}}}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle\int\int_{{{D}}}{0}{\left.{d}{x}\right.}{\left.{d}{y}\right.}={0}\) Therefore, the value of the line integrals is \(\displaystyle\oint_{{{C}}}{\frac{{{y}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}{\left.{d}{x}\right.}+{\frac{{-{\left({x}-{1}\right)}}}{{{\left({x}-{1}\right)}^{{{2}}}+{4}{y}^{{{2}}}}}}{\left.{d}{y}\right.}={0}\)