Question

Integrate the following:
$\int {\dfrac{{f\left( x \right)g'\left( x \right) - f'\left( x \right)g\left( x \right)}}{{f\left( x \right)g\left( x \right)}}\left[ {\log \left( {g\left( x \right)} \right) - \log \left( {f\left( x \right)} \right)} \right]dx = } $
A) $\log \left( {\dfrac{{g\left( x \right)}}{{f\left( x \right)}}} \right) + c$
B) $\dfrac{1}{2}{\left[ {\log \left( {\dfrac{{g\left( x \right)}}{{f\left( x \right)}}} \right)} \right]^2} + c$
C) $\dfrac{{\left( g \right)x}}{{f\left( x \right)}}\log \left( {\dfrac{{g\left( x \right)}}{{f\left( x \right)}}} \right) + c$
D) $\log \left[ {\dfrac{{g\left( x \right)}}{{f\left( x \right)}}} \right] - \dfrac{{g\left( x \right)}}{{f\left( x \right)}} + c$

Answer 1

Hint:
To solve this problem of integral, first we will use the properties of logarithm and then we will use substitution method of integration, where we substitute any part of a given function as an variable in the given function.

Complete step-by-step answer:
Given: we have given the integral $\smallint \dfrac{{f\left( x \right)g'\left( x \right) - f'\left( x \right)g\left( x \right)}}{{f\left( x \right)g\left( x \right)}}[\log \left( {g\left( x \right)} \right) - \log \left( {f\left( x \right)} \right)]dx$ and we have to find what it equals to.
So, at first let ${\text{I}}$ be equal to $\smallint \dfrac{{f\left( x \right)g'\left( x \right) - f'\left( x \right)g\left( x \right)}}{{f\left( x \right)g\left( x \right)}}[\log \left( {g\left( x \right)} \right) - \log \left( {f\left( x \right)} \right)]dx$
First we will take the log part and apply logarithm rules on it $\log \left( {g\left( x \right)} \right) - \log \left( {f\left( x \right)} \right)$
$ = \log \left( {\dfrac{{g\left( x \right)}}{{f\left( x \right)}}} \right)$
After putting the these values , we will get:
${\text{I}} = \int {\dfrac{{f\left( x \right)g'\left( x \right) - f'\left( x \right)g\left( x \right)}}{{f\left( x \right)g\left( x \right)}}} \left[ {\log \dfrac{{g\left( x \right)}}{{f\left( x \right)}}} \right]dx{\text{ }} \to \left( 1 \right)$
Now, let $\log \dfrac{{g\left( x \right)}}{{f\left( x \right)}} = t$
Now, we will apply partial differentiation method on it then we will get:
$\dfrac{1}{{\dfrac{{g\left( x \right)}}{{f\left( x \right)}}}}\left[ {\dfrac{{f\left( x \right)g'\left( x \right) - g\left( x \right)f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}}} \right]dx = dt$
Now, after simplifying this integral we will get:
$\dfrac{{f\left( x \right)g'\left( x \right) - g\left( x \right)f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}}dx = dt$
Now, after putting the values in equation (1) we will get:
${\text{I}} = \int {tdt} $
Now, integrate this term then we will get:
${\text{I}} = \dfrac{{{t^2}}}{2} + c$
We know, that the value of $t = \log \dfrac{{g\left( x \right)}}{{f\left( x \right)}}$
Now, substitute the value of $t$ in ${\text{I}} = \dfrac{{{t^2}}}{2} + c$
${\text{I}} = \dfrac{1}{2}\log {\left( {\dfrac{{g\left( x \right)}}{{f\left( x \right)}}} \right)^2} + c$
So, the value of integral of given term is $\dfrac{1}{2}\log {\left( {\dfrac{{g\left( x \right)}}{{f\left( x \right)}}} \right)^2} + c$.
Therefore, option B is the correct answer.

Note:
Never get confused during the use of property of $\log $ or logarithm rule. In this property:-
$\log x + \log y = \log xy$ and $\log x - \log y = \log \dfrac{x}{y}$.