Integrate $\ln \left( {2x} \right)$.

Hint:Indefinite integral simply represents the area under a given curve without any boundary conditions. So here by using this basic definition we can integrate $\ln \left( {2x} \right).$Also we know integration by parts: $\int {udv = uv - \int {vdu} } $.The above expression can also be used to integrate$\ln \left( {2x} \right)$.

Complete step by step answer:
Given, $\ln \left( {2x} \right)..............................\left( i \right)$
Also by the basic definition of indefinite integral we can write that:
Indefinite integral is given by: $\int {f\left( x \right)dx} $
Such to integrate $\ln \left( {2x} \right)$we can write
$\int {\ln \left( {2x} \right)dx} ..........................\left( {ii} \right)$
Now on observing (i) we can say that the term $\ln \left( {2x} \right)$cannot be integrated directly such that we have to use integration by parts, which is:
\[\int {udv = uv - \int {vdu} } ..................\left( {iii} \right)\]
Now here we need to find$u,du,v\;{\text{and}}\;dv$.
So from the given question we can write:
$u = \ln (2x)\;{\text{and}}\;dv = dx..................\left( {iv} \right)$
So we can find $v\;{\text{and}}\;du$from the given conditions in (iii):
$
u = \ln (2x)\;{\text{and}}\;dv = dx \\
\Rightarrow du = \dfrac{2}{{2x}} = \dfrac{1}{x}...............\left( v \right) \\
{\text{and}} \\
\Rightarrow v = x.......................\left( {vi} \right) \\ $
Now substituting (iv), (v) and (vi) in (iii), we get:
\[
\int {udv = uv - \int {vdu} } \\
\Rightarrow\int {\ln \left( {2x} \right)dx = uv - \int {vdu} } \\
\Rightarrow\int {\ln \left( {2x} \right)dx = \ln \left( {2x} \right) \times x - \int {\left( {x \times \dfrac{1}{x}} \right)dx} } \\
\Rightarrow\int {\ln \left( {2x} \right)dx = x\ln \left( {2x} \right) - \int {dx} } ..........................\left( {vii} \right) \\ \]
Now we know from basic integration identity:
$\int {dx = x} $
Substituting it in (vii) we get:
\[
\int {\ln \left( {2x} \right)dx = x\ln \left( {2x} \right) - \int {dx} } \\
\Rightarrow\int {\ln \left( {2x} \right)dx = x\ln \left( {2x} \right) - x} \\
\therefore\int {\ln \left( {2x} \right)dx = x\left( {\ln \left( {2x} \right) - 1} \right)} + C...........................\left( {viii} \right) \\
 \]
Therefore (viii) is our final answer.Since it’s an indefinite integral we have to add a constant C to the final answer.

So on integrating $\ln \left( {2x} \right)$ we get \[x\left( {\ln \left( {2x} \right) - 1} \right) + C.\]

Note:Since the basic definition indefinite integral simply implies the area under a curve such that the value of an integral must be finite or else the integral doesn’t exist. Similar to the above question if we can’t integrate an expression directly then we have to use the formula for integration by parts. Also common logarithmic functions are log functions with base 10, and natural logarithmic functions are log functions with base ‘e’. Natural logarithmic functions can be represented by$\ln x\;{\text{or}}\;{\log _e}x$.