Question:
$ \int{\frac{{{4}^{x+1}}-{{7}^{x-1}}}{{{28}^{x}}}}dx $ is equal to
$ \int{\frac{{{4}^{x+1}}-{{7}^{x-1}}}{{{28}^{x}}}}dx $ is equal to
Updated On: Jun 6, 2022
- $ \frac{1}{7{{\log }_{e}}4}{{4}^{-x}}-\frac{4}{{{\log }_{e}}7}{{7}^{-x}}+c $
- $ \frac{1}{7{{\log }_{e}}4}{{4}^{-x}}+\frac{4}{{{\log }_{e}}7}{{7}^{-x}}++c $
- $ \frac{{{4}^{-x}}}{{{\log }_{e}}7}-\frac{{{7}^{-x}}}{{{\log }_{e}}4}+c $
- $ \frac{{{4}^{-x}}}{{{\log }_{e}}4}-\frac{{{7}^{-x}}}{{{\log }_{e}}7}+c $
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The Correct Option is A
Solution and Explanation
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 $
$=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 $
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
