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) \[\frac{1}{28}{{\log }_{e}}{{4}^{-x}}+\frac{1}{7}{{\log }_{e}}{{7}^{-x}}+c\]
Correct Answer: A
Solution :
Let \[I=\int{\left\{ \frac{{{4}^{x+1}}}{{{28}^{x}}}-\frac{{{7}^{x-1}}}{{{28}^{x}}} \right\}}dx\] \[=4\int{\frac{1}{{{7}^{x}}}}dx-\frac{1}{7}\int{\frac{1}{{{4}^{x}}}}dx\] \[=-\frac{{{4.7}^{-x}}}{{{\log }_{e}}7}+\frac{1}{7{{\log }_{e}}4}{{4}^{-x}}+c\] \[=\frac{1}{7{{\log }_{e}}4}{{4}^{-x}}-\frac{{{4.7}^{-x}}}{{{\log }_{e}}7}+c\]You need to login to perform this action.
You will be redirected in
3 sec