Integration of a function provided as an implicit function Let a differentiable function $f$ satisfies the functional rule $$f(xy) = f(x) + f(y) + xy - x -y $$ for all values of $x,y > 0$ and $f'(1)=4$.
Based on this statement there were three questions:


*

*If $f(x_0)=0$, then $x_0$ lies in the interval
$$
(a) (0,1) \\
(b) (1,e) \\
(c) (e, e^2) \\
(d) (e^2,e^3)
$$ 

*$\int\frac{f(x)}{x}dx$ is equal to:
$$
(a) 3(\ln x)^2 + x + c \\
(b) 3(\ln x) + x + c \\
(c) 1.5(\ln x)^2 + x + c \\
(d) 1.5(\ln x) + x + c 
$$

*If $\int{e^{f(x)}}dx = e^x(ax^3 + bx^2 + cx + d) + \lambda$, then the value of $(a+b+c+d)$ is equal to:
$$
(a) -1 \\
(b) -2 \\
(c) 3 \\
(d) 6 
$$
My attempt : I had no idea how to deal with implicit function. Any hints or suggestion to handle this kind of question would have been extremely helpful. Thanks for giving me the pointers. I will try and update the full solutions soon.
 A: Differentiate $f(xy) = f(x) + f(y) + xy - x -y$ of the $y$ variable:
$$
xf'(xy)=f'(y)+x-1.
$$
Let $y=1$ then we get
$$
xf'(x)=f'(1)+x-1
$$
Since $f'(1)=4$ we get
$$
f'(x)=\frac{3}{x}+1
$$
Hence 
$$
f(x)=3\ln x +x+c
$$
where $c$ is an arbitrary constant.
Let $x=y=1$ in the original equation we get $f(1)=1$, hence we can get $c=0$, hence
$$
f(x)=3\ln x+x.
$$
Take this into the equation to examine it satisfies the condition.
Now you can solve the question directly.
A: The equation itself may be solved without any regularity conditions by observing that $g$, $g(x):=f(x)-x$ satisfies the logarithmic equation $g(xy)=g(x)+g(y)$ for $g\colon (0,\infty)\to\mathbb{R}$. Thus there is some additive function $a\colon\mathbb{R}\to\mathbb{R}$, i.e., $a$ satisfies $a(x+y)=a(x)+a(y)$ for all real $x,y$, such that $g(x)=a(\ln x)$ for all $x>0$. Thus $f$ with $f(x):=a(\ln x)+x$ is the general solutions. If $f$ is assumed to be regular, for example continuous, the function $a$ has to be of the form $x\mapsto c x$ with some constant $c$. Thus in particular holde true If $f$ is differentiable. In that case the condition on $f'$ results in the form of $f$ as given in the other answer.
