I'm having trouble figuring these out. If I have a function $\displaystyle f(x)=\frac{x^2+1}{x+1}$ and I want to find the oblique asymptote, where do I begin?
Divide $\displaystyle x+1$ into $\displaystyle x^2+1$
$\displaystyle
\begin{array}{c|ccc}
& x & - 1 & \\
\hline\\
x+1 & x^2 & 0 x & +1\\
& x^2 & x & 0\\
\hline\\ & & -x & +1\\
& & -x & -1\\
\hline\\
& & & 2 \end{array}
$
so $\displaystyle \frac{x^2+1}{x+1} = x -1 + \frac{2}{x+1}$ and your slant asymptote is $\displaystyle y = x - 1$ (as was mentioned).
What I did was I used polynomial long division and I obtained $\displaystyle f(x)=x+\frac{1-x}{1+x}$. If I use synthetic division:
[1] [1 0 1]
.........1 1
----------
.......1 1 2
So the remainder is $\displaystyle \frac{2}{x+1}$. I don't really understand where the $\displaystyle x-1$ comes from.
Did you do something like this:
$\displaystyle \frac{2}{x+1}+u=\frac{x^2+1}{x+1}$
Therefore $\displaystyle u$ is the slant asymptote, which turns out to be $\displaystyle x-1$