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Question

Mathematics Question on Integrals of Some Particular Functions

∫4x+1−7x−128xdx\int{\frac{{{4}^{x+1}}-{{7}^{x-1}}}{{{28}^{x}}}}dx is equal to

A

17log⁡e44−x−4log⁡e77−x+c\frac{1}{7{{\log }_{e}}4}{{4}^{-x}}-\frac{4}{{{\log }_{e}}7}{{7}^{-x}}+c

B

17log⁡e44−x+4log⁡e77−x++c\frac{1}{7{{\log }_{e}}4}{{4}^{-x}}+\frac{4}{{{\log }_{e}}7}{{7}^{-x}}++c

C

4−xlog⁡e7−7−xlog⁡e4+c\frac{{{4}^{-x}}}{{{\log }_{e}}7}-\frac{{{7}^{-x}}}{{{\log }_{e}}4}+c

D

4−xlog⁡e4−7−xlog⁡e7+c\frac{{{4}^{-x}}}{{{\log }_{e}}4}-\frac{{{7}^{-x}}}{{{\log }_{e}}7}+c

Answer

17log⁡e44−x−4log⁡e77−x+c\frac{1}{7{{\log }_{e}}4}{{4}^{-x}}-\frac{4}{{{\log }_{e}}7}{{7}^{-x}}+c

Explanation

Solution

Let I=\int{\left\\{ \frac{{{4}^{x+1}}}{{{28}^{x}}}-\frac{{{7}^{x-1}}}{{{28}^{x}}} \right\\}}dx
=4∫17xdx−17∫14xdx=4\int{\frac{1}{{{7}^{x}}}}dx-\frac{1}{7}\int{\frac{1}{{{4}^{x}}}}dx
=−4.7−xlog⁡e7+17log⁡e44−x+c=-\frac{{{4.7}^{-x}}}{{{\log }_{e}}7}+\frac{1}{7{{\log }_{e}}4}{{4}^{-x}}+c
=17log⁡e44−x−4.7−xlog⁡e7+c=\frac{1}{7{{\log }_{e}}4}{{4}^{-x}}-\frac{{{4.7}^{-x}}}{{{\log }_{e}}7}+c