Answer
$$\left( {\bf{a}} \right)\frac{{\ln \left( {{2^x} + 2} \right)}}{{\ln 2}} + C,\,\,\,\,\left( {\bf{b}} \right)\frac{1}{{\ln 2}}\ln \left( {43} \right)$$
Work Step by Step
$$\eqalign{
& \int_2^8 {\frac{{{2^x}}}{{{2^x} + 2}}} dx \cr
& \cr
& \left( {\bf{a}} \right){\text{write the general antiderivative}} \cr
& {\text{use substitution method}} \cr
& u = {2^x} + 2,\,\,\,\,du = {2^x}\ln 2dx,\,\,\,\,dx = \frac{{du}}{{{2^x}\ln 2}} \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{{2^x}}}{{{2^x} + 2}}} dx = \int {\frac{{{2^x}}}{u}} \left( {\frac{{du}}{{{2^x}\ln 2}}} \right) \cr
& = \int {\frac{1}{u}} \left( {\frac{{du}}{{\ln 2}}} \right) \cr
& = \frac{1}{{\ln 2}}\int {\frac{1}{u}} du \cr
& {\text{integrating}} \cr
& = \frac{1}{{\ln 2}}\ln \left| u \right| + C \cr
& {\text{write in terms of }}x \cr
& = \frac{{\ln \left| {{2^x} + 2} \right|}}{{\ln 2}} + C \cr
& = \frac{{\ln \left( {{2^x} + 2} \right)}}{{\ln 2}} + C \cr
& \cr
& \left( {\bf{b}} \right){\text{evaluate the definite integral}} \cr
& \int_2^8 {\frac{{{2^x}}}{{{2^x} + 2}}} dx = \left( {\frac{{\ln \left( {{2^x} + 2} \right)}}{{\ln 2}}} \right)_2^8 \cr
& = \frac{{\ln \left( {{2^8} + 2} \right)}}{{\ln 2}} - \frac{{\ln \left( {{2^2} + 2} \right)}}{{\ln 2}} \cr
& {\text{simplifying}} \cr
& = \frac{1}{{\ln 2}}\ln \left( {258} \right) - \frac{1}{{\ln 2}}\ln \left( 6 \right) \cr
& = \frac{1}{{\ln 2}}\left( {\ln \left( {258} \right) - \ln \left( 6 \right)} \right) \cr
& = \frac{1}{{\ln 2}}\ln \left( {\frac{{258}}{6}} \right) \cr
& = \frac{1}{{\ln 2}}\ln \left( {43} \right) \cr} $$