Question

Find the integral, $\int{\dfrac{({{4}^{x+1}}-{{7}^{x-1}})}{{{28}^{x}}}}=$
A. $\left( \dfrac{1}{7{{\log }_{e}}4} \right){{4}^{-x}}-\left( \dfrac{4}{{{\log }_{e}}7} \right){{7}^{-x}}+C$
B. $\left( \dfrac{1}{7{{\log }_{e}}4} \right){{4}^{-x}}+\left( \dfrac{4}{{{\log }_{e}}7} \right){{7}^{-x}}+C$
C. $\left( \dfrac{1}{{{\log }_{e}}7} \right)-\left( \dfrac{{{7}^{-x}}}{{{\log }_{e}}4} \right)+C$
D. $\left( \dfrac{{{4}^{-x}}}{{{\log }_{e}}4} \right)-\left( \dfrac{{{7}^{-x}}}{{{\log }_{e}}7} \right)+C$
E. $\left( \dfrac{1}{28} \right){{\log }_{e}}{{4}^{-x}}+\left( \dfrac{1}{7} \right){{\log }_{e}}{{7}^{-x}}+C$

Answer 1

Hint: In this question, we have to find the integration for which we should have knowledge of integration of ${{a}^{x}}$. So, simplify the terms and use the integration formula separately for each term. In order to do this, firstly we need to simplify the numerator and denominators for a simple form if possible. Then it becomes easy for us to solve the integration with appropriate formulae.

Formula Used: The formula used here is $\int{{{a}^{x}}}dx=\dfrac{{{a}^{x}}}{\log a}+c$ for $a\ne 1,a>0$
Here, formula used is valid in which $a\ne 1$ otherwise $\log a$ will become 0 and the integrated product will go to infinity.

Complete step by step solution:Firstly, separate the terms and solve in its simplest form before proceeding to integration.
$\begin{align}
  & \dfrac{{{4}^{x+1}}}{{{28}^{x}}}=\dfrac{{{4}^{x}}\times 4}{{{4}^{x}}\times {{7}^{x}}}=\dfrac{4}{{{7}^{x}}}\text{ }...(1) \\
 & \dfrac{{{7}^{x-1}}}{{{28}^{x}}}=\dfrac{{{7}^{x}}\times {{7}^{-1}}}{{{4}^{x}}\times {{7}^{x}}}=\dfrac{1}{7\times {{4}^{x}}}\text{ }...(2) \\
\end{align}$
Here, we have made them in their simplest form. Now, integrate the individual terms taking constants out of integration, i.e., 4 in the first term and 7 in the second term.
So,
$\begin{align}
  & I=\int{\left( \dfrac{{{4}^{x+1}}}{{{28}^{x}}}-\dfrac{{{7}^{x-1}}}{{{28}^{x}}} \right)dx} \\
 & \text{ }=\int{\left( \dfrac{4}{{{7}^{x}}}-\dfrac{1}{7\times {{4}^{x}}} \right)}dx \\
 & \text{ }=4\int{\dfrac{1}{{{7}^{x}}}dx-\dfrac{1}{7}}\int{\dfrac{1}{{{4}^{x}}}}dx \\
 & \text{ }=\dfrac{4\times {{7}^{-x}}}{-{{\log }_{e}}7}-\dfrac{{{4}^{-x}}}{-7{{\log }_{e}}4}+C \\
 & \text{ =}\dfrac{{{4}^{-x}}}{7{{\log }_{e}}4}-\dfrac{4\times {{7}^{-x}}}{{{\log }_{e}}7}+C \\
\end{align}$
Therefore, the given integral is $\int{\dfrac{({{4}^{x+1}}-{{7}^{x-1}})}{{{28}^{x}}}}=\left( \dfrac{1}{7{{\log }_{e}}4} \right){{4}^{-x}}-\left( \dfrac{4}{{{\log }_{e}}7} \right){{7}^{-x}}+C$


So, Option ‘A’ is correct

Note:This type question, we may get confused at negative signs and negative powers. Be sure with them and use appropriate formulae at the required place.