Question

$\int \frac{\left(4^{x+1}-7^{x-1}\right)}{{28}^x}=$

• A. $\left(\frac{1}{7{\log }_{e}4}\right)4^{-x}-\left(\frac{4}{{\log }_{e}7}\right)7^{-x}+C$
• B. $\left(\frac{1}{7{\log }_{e}4}\right)4^{-x}+\left(\frac{4}{{\log }_{e}7}\right)7^{-x}+C$
• C. $\left(\frac{1}{{\log }_{e}7}\right)-\left(\frac{7^{-x}}{{\log }_{e}4}\right)+C$
• D. $\left(\frac{4^{-x}}{{\log }_{e}4}\right)-\left(\frac{7^{-x}}{{\log }_{e}7}\right)+C$
• E. $\left(\frac{1}{28}\right){\log }_e{4}^{-x}+\left(\frac{1}{7}\right){\log }_e{7}^{-x}+C$

Answer 1

The correct option is A

$\left(\frac{1}{7{\log }_{e}4}\right)4^{-x}-\left(\frac{4}{{\log }_{e}7}\right)7^{-x}+C$

Explanation for the Correct answer:

Finding the value for the given integral:

$$\begin{gathered} \int \frac{\left(4^{x+1}-7^{x-1}\right)}{{28}^x}=\int \frac{4^{x+1}}{{28}^x}-\frac{7^{x-1}}{{28}^x}dx \\ =\int \frac{4^x.4}{{\left(4\times 7\right)}^x}-\frac{7^{x-1}}{{\left(4\times 7\right)}^x}dx \\ =\int \frac{4^x.4}{4^{{}^x}\times 7^{{}^x}}-\frac{7^x.7^{-1}}{4^{{}^x}\times 7^{{}^x}}dx \\ =\int \frac{4}{7^{{}^x}}dx-\int \frac{7^{-1}}{4^{{}^x}}dx \\ =4\int 7^{-x}dx-\frac{1}{7}\int 4^{{}^{-x}}dx \\ =4\left(-7^{{}^{-x}}{\log }_{e}7\right)-\frac{1}{7}\left(-4^{{}^{-x}}{\log }_{e}4\right)+c\left[\because \frac{d}{dx}{a}^x=a^x{\log }_{e}a\right] \\ =\frac{1}{7}\left(\frac{4^{-x}}{{\log }_{e}4}\right)-4\left(\frac{7^{-x}}{{\log }_{e}7}\right)+C\left[\because -{\log }_{e}a=\frac{1}{{\log }_{e}a}\right] \end{gathered}$$

Hence, option (A) is the correct answer.