Question

$ \int{\frac{{{4}^{x+1}}-{{7}^{x-1}}}{{{28}^{x}}}}dx $ is equal to

• A. $ \frac{1}{7{{\log }_{e}}4}{{4}^{-x}}-\frac{4}{{{\log }_{e}}7}{{7}^{-x}}+c $
• B. $ \frac{1}{7{{\log }_{e}}4}{{4}^{-x}}+\frac{4}{{{\log }_{e}}7}{{7}^{-x}}++c $
• C. $ \frac{{{4}^{-x}}}{{{\log }_{e}}7}-\frac{{{7}^{-x}}}{{{\log }_{e}}4}+c $
• D. $ \frac{{{4}^{-x}}}{{{\log }_{e}}4}-\frac{{{7}^{-x}}}{{{\log }_{e}}7}+c $
• E. $ \frac{1}{28}{{\log }_{e}}{{4}^{-x}}+\frac{1}{7}{{\log }_{e}}{{7}^{-x}}+c $

Answer 1

Answer: (A)

Let $ I=\int{\left\{ \frac{{{4}^{x+1}}}{{{28}^{x}}}-\frac{{{7}^{x-1}}}{{{28}^{x}}} \right\}}dx $
$=4\int{\frac{1}{{{7}^{x}}}}dx-\frac{1}{7}\int{\frac{1}{{{4}^{x}}}}dx $
$=-\frac{{{4.7}^{-x}}}{{{\log }_{e}}7}+\frac{1}{7{{\log }_{e}}4}{{4}^{-x}}+c $
$=\frac{1}{7{{\log }_{e}}4}{{4}^{-x}}-\frac{{{4.7}^{-x}}}{{{\log }_{e}}7}+c $