Answer
$$\frac{4}{{\ln 5}} - \frac{2}{{\ln 3}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\left( {{5^x} - {3^x}} \right)} dx \cr
& {\text{Integrate using the formula }}\int {{a^x}dx} = \frac{{{a^x}}}{{\ln a}} + C,{\text{ then}} \cr
& \int_0^1 {\left( {{5^x} - {3^x}} \right)} dx = \left( {\frac{{{5^x}}}{{\ln 5}} - \frac{{{3^x}}}{{\ln 3}}} \right)_0^1 \cr
& {\text{Evaluate the limits}} \cr
& = \left( {\frac{{{5^1}}}{{\ln 5}} - \frac{{{3^1}}}{{\ln 3}}} \right) - \left( {\frac{{{5^0}}}{{\ln 5}} - \frac{{{3^0}}}{{\ln 3}}} \right) \cr
& {\text{Simplify}} \cr
& = \frac{5}{{\ln 5}} - \frac{3}{{\ln 3}} - \frac{1}{{\ln 5}} + \frac{1}{{\ln 3}} \cr
& = \frac{{5 - 1}}{{\ln 5}} + \frac{{ - 3 + 1}}{{\ln 3}} \cr
& = \frac{4}{{\ln 5}} - \frac{2}{{\ln 3}} \cr} $$
