Answer
$+\infty$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to +\infty} [x-\ln (x^2+1)]$
Let us consider that $y=x-\ln (x^2+1) \implies e^y=\dfrac{e^x}{x^2+1}$
But $\lim\limits_{x \to +\infty} \dfrac{e^x}{x^2+1}=\dfrac{\infty}{\infty}$
We can see that the numerator and denominator have a limit of $\infty$, so the limit shows the indeterminate form of type $\dfrac{\infty}{\infty}$. So, we will apply L'Hopital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to +\infty} \dfrac{e^x}{x^2+1}=\lim\limits_{x \to +\infty} \dfrac{e^x}{2x}=+ \infty$
Thus, we have $\lim\limits_{x \to +\infty} y=\ln(+\infty)=+\infty$