Answer
$$\frac{{{2^x}}}{{\ln 2}} + \frac{{{5^x}}}{{\ln 5}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{4^x} + {{10}^x}}}{{{2^x}}}} dx \cr
& {\text{Distribute the numerator}} \cr
& = \int {\left( {\frac{{{4^x}}}{{{2^x}}} + \frac{{{{10}^x}}}{{{2^x}}}} \right)} dx \cr
& {\text{Rewrite using properties of exponents}} \cr
& = \int {\left( {\frac{{{{\left( {{2^2}} \right)}^x}}}{{{2^x}}} + \frac{{{{\left( {5 \cdot 2} \right)}^x}}}{{{2^x}}}} \right)} dx \cr
& = \int {\left( {\frac{{{2^{2x}}}}{{{2^x}}} + \frac{{\left( {{5^x}} \right)\left( {{2^x}} \right)}}{{{2^x}}}} \right)} dx \cr
& = \int {\left( {{2^x} + {5^x}} \right)} dx \cr
& = \int {{2^x}} dx + \int {{5^x}} dx \cr
& {\text{Integrating}} \cr
& = \frac{{{2^x}}}{{\ln 2}} + \frac{{{5^x}}}{{\ln 5}} + C \cr} $$
