$\int \frac{f(x) g'(x)+f'(x) g(x)}{f(x) g(x)}\{\log f(x)+\log g(x)\} d x$ is equal to
Answer & explanation
Correct answer: option 2
Let
$I =\int \frac{f(x) g'(x)+f'(x) g(x)}{f(x) g(x)}\{\log f(x)+\log g(x)\} d x$
$\Rightarrow I =\int \log \{f(x) g(x)\} \times \frac{1}{f(x) g(x)} d\{f(x) g(x)\}$
$\Rightarrow I =\int \log \{f(x) g(x)\} d[\log \{f(x) g(x)\}]$
$\Rightarrow I=\frac{1}{2}[\log \{f(x) g(x)\}]^2+C$