Question: Assume the sequence of Borel measurable functions $f_n:(0,1)\rightarrow\mathbb{R}$ satisfies $f_n(x)\geq\frac{1}{n}$ and $f_n(x)\rightarrow\infty$ as $n\rightarrow\infty$ for every $x\in(0,1)$. Is it necessarily true that $\int_0^1\frac{1}{(x+f_n(x))\log(n+1)}dx\rightarrow0$ as $n\rightarrow\infty$?
My Attempt: I was thinking of trying to use DCT. I suppose we could try and bound the integrand above by $\frac{1}{x+f_n(x)}$ since $\log(n+1)$ is strictly increasing as $n\rightarrow\infty$. Then, we know that, since $f_n(x)\geq\frac{1}{n}$, we have that $\frac{1}{x+f_n(x)}\leq\frac{1}{x+\frac{1}{n}}$, so using this as an integrable upper bound, we can pass the limit as $n\rightarrow\infty$ inside the integral and consequently we have $\int_0^1\frac{1}{x}dx=\ln|x|]_0^1$, which diverges. So, it isn't necessarily true that the integral goes to $0$ as $n\rightarrow\infty$. I suppose I am just not sure if I did this right, or if I made a mistake somewhere. Any thoughts, ideas, different approaches, etc. are greatly appreciated! Thank you.
If $a_n\in(0,1)$ form a decreasing sequence with $\lim\limits_{n\to\infty}a_n=0$, then $$f_n(x)=\begin{cases}1/n,&0<x<a_n\\n,&a_n\leqslant x<1\end{cases}$$ satisfies the premises, and $$\int_0^1\frac{dx}{x+f_n(x)}\geqslant\int_0^{a_n}\frac{dx}{x+f_n(x)}=\log(1+na_n).$$ So, for a counterexample, it suffices to have $\log(1+na_n)/\log(1+n)\not\to 0$ as $n\to\infty$.
In fact, with $a_n=1/\log(n+1)$ (say), the limit takes its greatest possible value $1$.