How do you integrate \[\ln (5x + 3)?\]

Hint: We will use the by parts formula of integration to solve the given integration. On doing some simplification we get the required answer.
The by parts formula of integration is as following:
Let's say, we are integrating two terms \[u,v\]with the respect of \[x\].
So, we can derive the formula as following:
\[\int {uvdx} = u\int {vdx - \int {\left[ {\dfrac{{du}}{{dx}}\int {vdx} } \right]} } dx\].
But which term will consider as “\[u\]”, depends on the following rule:
LIATE:
‘L’ stands for Logarithm.
‘I’ stands for Inverse.
‘A’ stands for Algebra.
‘T’ stands for Trigonometric.
‘E’ stands for exponential.
We will also use the following integration formulas:
\[\dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x} + K\], where \[K\]is an arbitrary constant.

Complete step by step answer:
The following integration has to be integrating: \[\int {\ln (5x + 3)} .dx\]
So, By applying the by parts rule, we can consider the following term as \[u\] and \[v\].
Let's say,\[(5x + 3) = u\].
So, taking the derivation of the first term, \[u\], with respect to \[x\], we get:
\[ \Rightarrow \dfrac{{du}}{{dx}} = 5\].
So, we can re-write it as following:
\[ \Rightarrow du = 5dx\].
So, putting the above integration in by parts formula, we can write it as following:
\[ \Rightarrow \dfrac{1}{5}\int {\ln (u).du} \]
By solving it, we get: \[\dfrac{1}{5}\int {\ln (u)du} \]
\[ \Rightarrow \dfrac{1}{5}\left[ {\ln (u)\int {1du} - \int {\left[ {\dfrac{d}{{du}}(\ln (u))\int {1du} } \right]du} } \right]\]
Let us simplify the term and we get,
\[ \Rightarrow \dfrac{1}{5}\left[ {u\ln (u) - \int {\dfrac{1}{u}udu} } \right]\]
On simplify we get
\[ \Rightarrow \dfrac{1}{5}\left[ {u\ln (u) - \int {du} } \right]\]
Then we get,
\[ \Rightarrow \dfrac{1}{5}\left[ {u\ln (u) - u} \right] + K\]
Now, putting the values of \[u,v\], we get the following iteration:
\[\int {\ln (5x + 3)dx} = \dfrac{1}{5}\left[ {(5x + 3)\ln (5x + 3) - (5x + 3)} \right] + K\], where \[K\]is an arbitrary constant.

Therefore, the answer of the above question is ”\[\dfrac{{(5x + 3)}}{5}\left[ {\ln (5x + 3) - 1} \right] + K\], where \[K\] is an arbitrary constant”.

Note: Points to remember: As there is no direct approach present in mathematics to solve integration of logarithms, we need to apply by parts rule in such cases. Priority of the terms will follow the rule of “LIATE”.