Let f(xy) = f(x) f(y) for all x > 0, y > 0 and f(1 + x) = 1 + x {1 + g(x)}, where $\lim\limits_{x \rightarrow 0} g(x)=0$, then $\int \frac{f(x)}{f'(x)} d x$, is

$\frac{x^2}{2}+C$

We have,

f(xy) = f(x) f(y) for all x > 0, y > 0

$\Rightarrow f(x)=x^n$

$\Rightarrow f(1+x)=(1+x)^n$

$\Rightarrow 1+x\{1+g(x)\}=(1+x)^n$

$\Rightarrow 1+x\{1+g(x)\}=1+n x$       [∵ (1+x)ⁿ ≃ 1+n x, if x → 0]

$\Rightarrow 1+g(x)=n$ for all x in the neighbourhood of x = 0

$\Rightarrow g(x)=n-1$ for all x in the neighbourhood of x = 0

But, $\lim\limits_{x \rightarrow 0} g(x)=0$

∴  $n-1=0 \Rightarrow n=1$

Hence, f(x) = x

∴  $\int \frac{f(x)}{f'(x)} d x=\int x d x=\frac{x^2}{2}+C$