Let $f(x):=\frac{x^2}{1+x}$ for $x\geq 0$.
Function $f$ is continuous in $[0,\infty[$ and its graph has an oblique asymptote when $x$ tends to $\infty$ (the asymptote being the straight line of equation $y=x-1$). Finally you can conclude using the following lemma:
Let $a\in \mathbb{R}$ and $f:[a,\infty[\to \mathbb{R}$ a continuous function.
If the graph of $f$ has a horizontal or an oblique asymptote when $x$ tends to $\infty$, then $f$ is uniformly continuous in $[a,\infty[$.
The lemma cited above is a corollary of the following theorem:
Let $a\in \mathbb{R}$ and $f,g:[a,\infty[\to \mathbb{R}$ be continuous functions such that: $\tag{A} \lim_{x\to \infty} f(x)-g(x)=0\; .$ Then $f$ is uniformly continuous in $[a,\infty[$ if and only if $g$ is uniformly continuous in $[a,\infty[$.
The proof is very simple.
Since the statement is symmetric w.r.t. $f$ and $g$, it suffices to prove only one implication; therefore we assume, e.g., that $g$ is u.c. in $[a,\infty[$ and prove that also $f$ is u.c. in $[a,\infty[$. Moreover, w.l.o.g., we assume also $a=0$.
Let $\varepsilon >0$ be fixed. By (A) we can find $r\geq 0$ such that: $\tag{I} \forall x\geq r,\quad |f(x)-g(x)|\leq \frac{\varepsilon}{8}\; ;$ since $g$ is u.c. in $[0,\infty[$ we can find $\delta_1>0$ such that: $\tag{II} \forall x,y\geq 0,\quad |x-y|\leq \delta_1\ \Rightarrow\ |g(x)-g(y)|\leq \frac{\varepsilon}{4}\; ;$ finally, by Heine-Cantor theorem, we can find $\delta_2>0$ such that: $\tag{III} \forall 0\leq x,y\leq r,\quad |x-y|\leq \delta_2\ \Rightarrow\ |f(x)-f(y)|\leq \frac{\varepsilon}{2}\; .$ Now, set $\delta:=\min \{\delta_1, \delta_2\}$ and take two arbitrary $x,y\geq 0$ s.t. $|x-y|\leq \delta$: then, due to the symmetry w.r.t. $x,y$, three cases arise:
$r\leq x,y$: in this case we find: $|f(x)-f(y)|\leq |f(x)-g(x)|+|g(x)-g(y)|+|g(y)-f(y)|\; ,$ hence $|f(x)-f(y)|\leq \frac{\varepsilon}{8}+\frac{\varepsilon}{4}+\frac{\varepsilon}{8}=\frac{\varepsilon}{2}\leq \varepsilon$ because of (I) & (II)
$0\leq x,y\leq r$: in such a case $|f(x)-f(y)|\leq \frac{\varepsilon}{2}\leq \varepsilon$ by (III);
$0\leq x\leq r < y$: in this case we have: $|f(x)-f(y)|\leq |f(x)-f(r)| +|f(r)-f(y)|\; ,$ thus $|f(x)-f(y)|\leq \frac{\varepsilon}{2} +\frac{\varepsilon}{2}=\varepsilon$ for (III) & case 1.
Therefore $|x-y|\leq \delta\ \Rightarrow\ |f(x)-f(y)|\leq \varepsilon$ for all $x,y\geq 0$ and $f$ is u.c. in $[0,\infty[$. $\square$