Let $f$ be a function such that $f(0)=0$ and $|f'(x)|\le\frac1{1+x}$ for $x\ge0.$ Prove that $\int_0^{e-1}\big(f(x)\big)^2\,dx\le e-2.$
2. $\left| f(x) \right|=\left| \int_{0}^{x}{f'(t)\,dt} \right|\le \int_{0}^{x}{\left| f'(t) \right|\,dt}\le \int_{0}^{x}{\frac{dt}{1+t}}=\ln (1+x).$
$\int_{0}^{e-1}{{{\left| f(x) \right|}^{2}}\,dx}\le \int_{0}^{e-1}{{{\ln }^{2}}(1+x)\,dx}=e-2.\quad\blacksquare$