Answer
$$A = \frac{5}{2}\ln \left( {4 + \sqrt {17} } \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{{5x}}{{\sqrt {{x^4} + 1} }} \cr
& {\text{The area of the region is given by }} \cr
& A = \int_0^2 {\frac{{5x}}{{\sqrt {{x^4} + 1} }}} dx \cr
& A = \frac{5}{2}\int_0^2 {\frac{{2x}}{{\sqrt {{{\left( {{x^2}} \right)}^2} + 1} }}} dx \cr
& {\text{Using theorem 5}}{\text{.20 }}\int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }} = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C} \cr
& A = \frac{5}{2}\left[ {\ln \left( {{x^2} + \sqrt {{{\left( {{x^2}} \right)}^2} + 1} } \right)} \right]_0^2 \cr
& A = \frac{5}{2}\left[ {\ln \left( {{2^2} + \sqrt {{{\left( {{2^2}} \right)}^2} + 1} } \right) - \ln \left( {{0^2} + \sqrt {{{\left( 0 \right)}^2} + 1} } \right)} \right] \cr
& A = \frac{5}{2}\left[ {\ln \left( {4 + \sqrt {17} } \right) - \ln \left( 1 \right)} \right] \cr
& A = \frac{5}{2}\ln \left( {4 + \sqrt {17} } \right) \cr} $$
