Answer
$$A = 2\sqrt 2 + 2\ln \left( {1 + \sqrt 2 } \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {\left( {4 + {x^2}} \right)^{1/2}},\,\,\,\,\left[ {0,2} \right] \cr
& {\text{The area under de curve is given by}} \cr
& A = \int_0^2 {{{\left( {4 + {x^2}} \right)}^{1/2}}} dx \cr
& {\text{The integrand contains the form }}{a^2} + {x^2} \cr
& 4 + {x^2} \to a = 2 \cr
& {\text{Use the change of variable }}x = a\tan \theta \cr
& x = 2\tan \theta ,\,\,\,dx = 2{\sec ^2}\theta d\theta \cr
& = \int {{{\left( {4 + 4{{\tan }^2}\theta } \right)}^{1/2}}\left( {2{{\sec }^2}\theta } \right)d\theta } \cr
& = 4\int {{{\left( {1 + {{\tan }^2}\theta } \right)}^{1/2}}{{\sec }^2}\theta d\theta } \cr
& = 4\int {{{\left( {{{\sec }^2}\theta } \right)}^{1/2}}{{\sec }^2}\theta d\theta } \cr
& = 4\int {{{\sec }^3}\theta d\theta } \cr
& {\text{Integrate using the reduction formula}} \cr
& = 4\left[ {\frac{1}{2}\left( {\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right|} \right)} \right] + C \cr
& = 2\sec \theta \tan \theta + 2\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{Write in terms of }}x \cr
& = 2\left( {\frac{{\sqrt {4 + {x^2}} }}{2}} \right)\left( {\frac{x}{2}} \right) + 2\ln \left| {\frac{{\sqrt {4 + {x^2}} }}{2} + \frac{x}{2}} \right| + C \cr
& = \frac{{x\sqrt {4 + {x^2}} }}{2} + 2\ln \left| {\frac{{\sqrt {4 + {x^2}} + x}}{2}} \right| + C \cr
& \cr
& ,{\text{then}} \cr
& A = \int_0^2 {{{\left( {4 + {x^2}} \right)}^{1/2}}} dx \cr
& A = \left( {\frac{{2\sqrt {4 + {2^2}} }}{2} + 2\ln \left| {\frac{{\sqrt {4 + {2^2}} + 2}}{2}} \right|} \right) - \left( {\frac{0}{2} + 2\ln \left| {\frac{{\sqrt {4 + {0^2}} + 0}}{2}} \right|} \right) \cr
& A = 2\sqrt 2 + 2\ln \left| {1 + \sqrt 2 } \right| - 2\ln \left| 1 \right| \cr
& A = 2\sqrt 2 + 2\ln \left( {1 + \sqrt 2 } \right) \cr} $$
