Answer
$$Area = 4\sqrt 2 + \frac{1}{3}\ln \left( {\frac{{3 - 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right)$$
Work Step by Step
$$\eqalign{
& y = 0\left( {x{\text{ - axis}}} \right),{\text{ }}y = \frac{{{x^2} - 4x - 4}}{{{x^2} - 4x - 5}} \cr
& {\text{setting }}\frac{{{x^2} - 4x - 4}}{{{x^2} - 4x - 5}} = 0 \cr
& {x^2} - 4x - 4 = 0 \cr
& {\text{quadratic formula}} \cr
& x = \frac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4\left( 1 \right)\left( { - 4} \right)} }}{2} \cr
& x = \frac{{4 \pm \sqrt {32} }}{2} \cr
& x = \frac{{4 \pm 4\sqrt 2 }}{2} \cr
& x = 2 \pm 2\sqrt 2 \cr
& {\text{interval }}\left[ {2 - 2\sqrt 2 ,2 + 2\sqrt 2 } \right] \cr
& \frac{{{x^2} - 4x - 4}}{{{x^2} - 4x - 5}} \geqslant 0{\text{ in the interval }}\left[ {2 - 2\sqrt 2 ,2 + 2\sqrt 2 } \right] \cr
& {\text{the area of the region is}} \cr
& Area = \int_{2 - 2\sqrt 2 }^{2 + 2\sqrt 2 } {\left( {\frac{{{x^2} - 4x - 4}}{{{x^2} - 4x - 5}}} \right)} dx \cr
& {\text{use long division}} \cr
& Area = \int_{2 - 2\sqrt 2 }^{2 + 2\sqrt 2 } {\left( {1 + \frac{1}{{{x^2} - 4x - 5}}} \right)} dx \cr
& \frac{1}{{{x^2} - 4x - 5}} = \frac{1}{{\left( {x - 5} \right)\left( {x + 1} \right)}} \cr
& {\text{partial fractions}} \cr
& \frac{1}{{\left( {x - 4} \right)\left( {x + 1} \right)}} = \frac{A}{{x - 5}} + \frac{B}{{x + 1}} \cr
& {\text{multiplying}} \cr
& 1 = A\left( {x + 1} \right) + B\left( {x - 5} \right) \cr
& {\text{Letting }}x = 5 \cr
& 1 = A\left( 6 \right) + B\left( 0 \right) \cr
& A = \frac{1}{6} \cr
& {\text{Letting }}x = - 1 \cr
& 1 = A\left( 0 \right) + B\left( { - 6} \right) \cr
& B = - \frac{1}{6} \cr
& {\text{substituting constants}} \cr
& \frac{1}{{\left( {x - 4} \right)\left( {x + 1} \right)}} = \frac{1}{{6\left( {x - 5} \right)}} - \frac{1}{{6\left( {x + 1} \right)}} \cr
& Area = \int_{2 - 2\sqrt 2 }^{2 + 2\sqrt 2 } {\left( {1 + \frac{1}{{{x^2} - 4x - 5}}} \right)} dx = \int_{2 - 2\sqrt 2 }^{2 + 2\sqrt 2 } {\left( {1 + \frac{1}{{6\left( {x - 5} \right)}} - \frac{1}{{6\left( {x + 1} \right)}}} \right)} dx \cr
& {\text{integrate}} \cr
& Area = \left( {\left[ {x + \frac{1}{6}\ln \left| {x - 5} \right| - \frac{1}{6}\ln \left| {x + 1} \right|} \right]} \right)_{2 - 2\sqrt 2 }^{2 + 2\sqrt 2 } \cr
& {\text{evaluate limits}} \cr
& Area = \left[ {2 + 2\sqrt 2 + \frac{1}{6}\ln \left| {2 + 2\sqrt 2 - 5} \right| - \frac{1}{6}\ln \left| {2 + 2\sqrt 2 + 1} \right|} \right] \cr
& - \left[ {2 - 2\sqrt 2 + \frac{1}{6}\ln \left| {2 - 2\sqrt 2 - 5} \right| - \frac{1}{6}\ln \left| {2 - 2\sqrt 2 + 1} \right|} \right] \cr
& Area = \left[ {2 + 2\sqrt 2 + \frac{1}{6}\ln \left| {2\sqrt 2 - 3} \right| - \frac{1}{6}\ln \left| {3 + 2\sqrt 2 } \right|} \right] \cr
& - \left[ {2 - 2\sqrt 2 + \frac{1}{6}\ln \left| { - 3 - 2\sqrt 2 } \right| - \frac{1}{6}\ln \left| {3 - 2\sqrt 2 } \right|} \right] \cr
& simplifying \cr
& Area = \left[ {2 + 2\sqrt 2 + \frac{1}{6}\ln \left| {\frac{{2\sqrt 2 - 3}}{{2\sqrt 2 + 3}}} \right|} \right] - \left[ {2 - 2\sqrt 2 + \frac{1}{6}\ln \left| {\frac{{3 + 2\sqrt 2 }}{{3 - 2\sqrt 2 }}} \right|} \right] \cr
& Area = 2 + 2\sqrt 2 + \frac{1}{6}\ln \left| {\frac{{2\sqrt 2 - 3}}{{2\sqrt 2 + 3}}} \right| - 2 + 2\sqrt 2 - \frac{1}{6}\ln \left| {\frac{{3 + 2\sqrt 2 }}{{3 - 2\sqrt 2 }}} \right| \cr
& Area = 4\sqrt 2 + \frac{1}{6}\ln \left| {\frac{{2\sqrt 2 - 3}}{{2\sqrt 2 + 3}}} \right| + \frac{1}{6}\ln \left| {\frac{{3 - 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right| \cr
& Area = 4\sqrt 2 + \frac{1}{3}\ln \left( {\frac{{3 - 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right) \cr} $$
