Answer
$y = x-1~~$ is a slant asymptote.
Work Step by Step
$y = \frac{x^2+1}{x+1}$
We can express $x^2+1$ in terms of $x+1$:
$x^2+1 = (x+1)(x-1)+2$
Then:
$\frac{x^2+1}{x+1} = (x-1)+\frac{2}{x+1}$
We can see that $~~y = x-1~~$ is a good candidate for a slant asymptote.
We can evaluate the limit as $x \to -\infty$:
$\lim\limits_{x \to -\infty} [\frac{x^2+1}{x+1}-(x-1)]$
$= \lim\limits_{x \to -\infty} [\frac{x^2+1}{x+1}-\frac{(x-1)(x+1)}{x+1}]$
$= \lim\limits_{x \to -\infty} (\frac{x^2+1}{x+1}-\frac{x^2-1}{x+1})$
$= \lim\limits_{x \to -\infty} \frac{2}{x+1}$
$= 0$
We can evaluate the limit as $x \to \infty$:
$\lim\limits_{x \to -\infty} [\frac{x^2+1}{x+1}-(x-1)]$
$= \lim\limits_{x \to -\infty} [\frac{x^2+1}{x+1}-\frac{(x-1)(x+1)}{x+1}]$
$= \lim\limits_{x \to -\infty} (\frac{x^2+1}{x+1}-\frac{x^2-1}{x+1})$
$= \lim\limits_{x \to -\infty} \frac{2}{x+1}$
$= 0$
Therefore, $~~y = x-1~~$ is a slant asymptote.