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