Is $f(x) = \log (1 +x^2) $ uniformly continious on $\mathbb{R}$? Yes/No.
My attempt : yes
Let $\Delta y:=x-y$, so that $x=y+\Delta y$. Then $$ \frac{1+x^2}{1+y^2}=\frac{1+y^2+2y\Delta y+(\Delta y)^2}{1+y^2}=1+\frac{2y\Delta y}{1+y^2}+\frac{(\Delta y)^2}{1+y^2}. $$ From here, we can note that for all $y\in\mathbb{R}$, $$ \left\lvert\frac{2y}{1+y^2}\right\rvert\leq 1\qquad\text{and}\qquad\left\lvert\frac{1}{1+y^2}\right\rvert\leq 1. $$
(The second inequality is immediate; for the first try considering $\lvert y\rvert\leq 1$ and $\lvert y\rvert>1$ as separate cases.
But, as a result, $$ \left\lvert\frac{1+x^2}{1+y^2}-1\right\rvert\leq\left\lvert\frac{2y}{1+y^2}\right\rvert\lvert\Delta y\rvert+\left\lvert\frac{1}{1+y^2}\right\rvert(\Delta y)^2\leq\lvert\Delta y\rvert+(\Delta y)^2. $$ we can definitely make this last expression small by making $\Delta y$ small... and note that $\lvert\Delta y\rvert<\delta$.
You can observe that$$(\forall x\in\mathbb R):\bigl\lvert f'(x)\bigr\rvert=\left\lvert\frac{2x}{x^2+1}\right\rvert\leqslant1.$$It follows from this and from the Mean Value Theorem that$$\lvert x-y\rvert<\varepsilon\implies\bigl\lvert f(x)-f(y)\bigr\rvert<\varepsilon.$$