Question

What is the integral of $ln(2x)$?

Answer 1

Hint: We need to find the integral of $ln(2x)$. For this we need to perform by parts, because this is not any general form which can be integrated directly. Since there is only one term here, we need to plug in the second term by ourselves and it is obvious that the term would be 1. So, we convert $ln(2x)$ as the product of 1 and $ln(2x)$ and then apply by parts on it.

Complete step-by-step answer:
If we do not have the direct formula while integrating a function we apply a technique called ‘by parts’. We write $ln(2x)$ as $ln(2x)\times 1$ and then apply the formula below:
$\int{\text{ (}}a\left( x \right)\times b\left( x \right)dx)=a\left( x \right)\times \int{b}\left( x \right)dx-\int{\left( \dfrac{d\left( a\left( x \right) \right)}{dx}\times \int{b}\left( x \right) \right)dx}$
Now, the order in which the functions are to be taken is also chosen using a rule called “ILATE” where we give preference to the inverse functions first, then logarithmic and arithmetic, then trigonometric and last preference is given to exponential functions.
Here we take the first function to be $ln(2x)$ since it comes first on the list and then we take 1 to be the second function.
So here we put:
$a\left(x\right)=ln(2x)$
$b\left(x\right)=1$
$\int \left(ln(2x)\times 1 \right)dx=ln(2x)\times \int 1 dx+\int \left(\dfrac{2}{2x}\times x\right)dx$
We have used the formulae:
$\dfrac{ln(2x)}{dx}=2\times\dfrac{1}{2x}=\dfrac{1}{x}$
$\int 1dx=x$
Now, simplifying it further:
$\int \left(ln(2x)\times 1 \right)dx=ln(2x)\times x+\int \left(1\right)dx$
Solving it further, we have:
$\int \left(ln(2x)\times 1 \right)dx=x\times ln(2x)+x=x\left(ln(2x)+1\right)$
Hence, the integral is solved.

Note: The most frequent mistake while doing integration of functions involving a linear term inside some function is that we tend to differentiate the function with respect to the independent variable. For example, in this case the most common answer that would be obtained will be $\dfrac{2}{2x}$ which is the differential of the function not the integral. So be very aware while performing this operation.