Bounds of $\frac{\ln(x+1)}{x}\ \forall x>0$ $f:(0,\infty)$,
$f(x)=\frac{\ln(x+1)}{x}$
Prove that for $\forall\ x>0$ that $f(x) \in(0,1)$. I calculated the derivative of $f(x)$: $f'(x)=\frac{\frac{x}{x+1}-\ln(1+x)}{x^2}$ which I think simplifies to $\frac{x^3}{x+1}-x^2(\ln(1+x))$. I have no idea what to do next, I can't find the roots of this equation and I don't see any connection as of why it should be bounded by 0 and 1. 
I hope I formatted this well, I don't usually post here but I am really curious how could I solve this kind of exercise. 
 A: First of all if is easy to see that $\dfrac{ln{(x+1)}}{x}>0$.
It's well known that $\ln{x}\leq x-1$ where the equality is valid only for $x=1$. (You can prove it if you define the function $f(x)=\ln{x}-x+1$ and find the global maximum using simple calculus). Then by this we get that $\ln{(x+1)}\leq x$ so for $x>0$ we get $\dfrac{ln{(x+1)}}{x}<1$.
Alexandros
A: For $x \in (0,\infty)$, you have
$$\ln (x+1) = \int_0^x \frac{dt}{1+t}$$
Hence
$$0 \le f(x) = \frac{1}{x}\int_0^x \frac{dt}{1+t} < \frac{1}{x}\int_0^x \ dt =1$$
As all considered maps are continuous.
A: $\ln(x+1)$ is a concave function, so the curve is under each of it tangents. Now it happens that the line with equation $y=x$ is its tangent at origin. So for any $x 0$ of its domain, $\ln(x+1)\le x$. Furthermore, considering the function $x-\ln(x+1)$, it is easy to see this function is increasing on [0,+\infty), so
$$\ln(x+1)<x\enspace\forall x>0\iff \frac{\ln(x+1)}x<1\enspace\forall x>0.$$
The inequality $\dfrac{\ln(x+1)}x>0$ is obvious since $1+x>1$.
A: It is much simpler to show that the derivative of $\ln(x+1)$ is always in $(0,1)$ (and it is continuous). 
Then $\ln(x+1)$ is strictly between $0$ and $x$, think about it. 
