Answer
362.1k+ views
Hint: The equation in the question is the simple type of integration that can be solved using some simple operation.
We can solve it by adding a term or number in the both numerator or denominator.
As well as we can solve it by rationalizing the denominator method.
Complete step by step solution:
Let us take the equation,
$ \Rightarrow $ $ \int {\dfrac{{3x}}{{3x - 1}}dx} $
By adding and subtracting $ 1 $ in both numerator and denominator,
$ \Rightarrow $ $ \int {\dfrac{{\left( {3x - 1} \right) + 1}}{{3x - 1}}} $
$ \Rightarrow \int {1dx + \int {\dfrac{{dx}}{{3x - 1}}} } $
By integrating and using logarithmic integration,
$ \Rightarrow x + \dfrac{1}{3}\log (3x - 1) + c $
So, the correct answer is “ $ x + \dfrac{1}{3}\log (3x - 1) + c $ ”.
Note:
> Integration is the calculation of an integral.
> The integration denotes the summation of discrete data. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly.
We can solve it by adding a term or number in the both numerator or denominator.
As well as we can solve it by rationalizing the denominator method.
Complete step by step solution:
Let us take the equation,
$ \Rightarrow $ $ \int {\dfrac{{3x}}{{3x - 1}}dx} $
By adding and subtracting $ 1 $ in both numerator and denominator,
$ \Rightarrow $ $ \int {\dfrac{{\left( {3x - 1} \right) + 1}}{{3x - 1}}} $
$ \Rightarrow \int {1dx + \int {\dfrac{{dx}}{{3x - 1}}} } $
By integrating and using logarithmic integration,
$ \Rightarrow x + \dfrac{1}{3}\log (3x - 1) + c $
So, the correct answer is “ $ x + \dfrac{1}{3}\log (3x - 1) + c $ ”.
Note:
> Integration is the calculation of an integral.
> The integration denotes the summation of discrete data. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly.
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