How do you integrate \[\int {\ln 4xdx} \] using integration by parts?

Hint: We use the concept of integration by parts to integrate the given integral. Using the ILATE sequence names the first and second part and then substitutes the differentiations and parts in the formula.
*In integration by parts we use the sequence of ILATE (Inverse trigonometric, Logarithms, algebraic, trigonometric, exponential) for choosing the value of \[u,v\] and then substitute in the integration formula of by parts integration.
* If we have \[\int {uvdx} \] then the value of integral is given by \[\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)dx} } } \] where \[u' = \dfrac{{du}}{{dx}}\] is differentiation of \[u\] with respect to \[x\].

Complete step-by-step answer:
We have \[\int {\ln 4xdx} \]
Here there are two parts in the integration, one is x and one is logarithm 4
Now we will solve the integration by using integration by parts.
Looking at the integral \[\int {\ln 4xdx} \]
Using the ILATE sequence we can tell that the term comes first in the sequence is\[u\]and which comes later is \[v\] i.e. in this case we have the terms exponential and algebraic , first comes the logarithm term then comes the algebraic term so \[u\] is \[\ln 4\] and \[v\] is \[x\]
\[u = \ln 4,v = x\]
Now we calculate the differentiation of the term \[u\]
\[u' = \dfrac{{du}}{{dx}}\]
Substitute the value of u:
\[u' = \dfrac{{d(\ln 4)}}{{dx}}\]
\[u' = \dfrac{1}{x}\]
Now substitute the values in equation \[\int {uvdx = u\int {vdx - \int {u'\left( {\int {vdx} } \right)dx} } } \]
\[ \Rightarrow \int {\ln 4xdx} = \ln 4\int {xdx} - \int {\dfrac{1}{x}\left( {\int x dx} \right)dx} \] … (1)
Now we know \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\]
\[ \Rightarrow \int {\ln 4xdx} = \ln 4\left( {\dfrac{{{x^2}}}{2}} \right) - \int {\dfrac{1}{x}\left( {\dfrac{{{x^2}}}{2}} \right)dx} \]
Now we can solve and bring out constant terms from the integral sign.
\[ \Rightarrow \int {\ln 4xdx} = \ln 4\left( {\dfrac{{{x^2}}}{2}} \right) - \dfrac{1}{2}\int {\dfrac{{{x^2}}}{x}dx} \]
Cancel same terms from numerator and denominator wherever possible
\[ \Rightarrow \int {\ln 4xdx} = \ln 4\left( {\dfrac{{{x^2}}}{2}} \right) - \dfrac{1}{2}\int {xdx} \]
Again use the formula \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C\]
\[ \Rightarrow \int {\ln 4xdx} = \ln 4\left( {\dfrac{{{x^2}}}{2}} \right) - \dfrac{1}{2}\left( {\dfrac{{{x^2}}}{2}} \right) + C\]
Open brackets and multiply the terms
\[ \Rightarrow \int {\ln 4xdx} = \left( {\dfrac{{{x^2}}}{2}} \right)\ln 4 - \left( {\dfrac{{{x^2}}}{4}} \right) + C\]
Take common the value \[\left( {\dfrac{{{x^2}}}{2}} \right)\]
\[ \Rightarrow \int {\ln 4xdx} = \left( {\dfrac{{{x^2}}}{2}} \right)\left[ {\ln 4 - \dfrac{1}{2}} \right] + C\]

\[\therefore \]The integration of \[\int {\ln 4xdx} \] using integration by parts is \[\left( {\dfrac{{{x^2}}}{2}} \right)\left[ {\ln 4 - \dfrac{1}{2}} \right] + C\]

Note:
Students are likely to make the mistake of assuming the ‘I’ in ILATE as integer and then must write x as the first part in the integration by parts which is wrong. Keep in mind that ‘x’ is an algebraic term and I denote inverse function. Also, use integration of log directly in the solution.