For improper fraction $\left[\frac{p(x)}{q(x)}\right]$
(i) $P(x)$ and $q(x)$ both are polynomial.
(ii) Degree of $p(x) \geq$ degree of $q(x)$.
Option (a) $\frac{x^2+1}{\left(x^2+1\right)\left(x^2+x+1\right)}=\frac{p(x)}{q(x)}$
Degree of $p(x)=2$
Degree of $q(x)=4$
It is not improper fraction.
Option (b) $\frac{x^2+1}{(x+3)\left(x^2-x+1\right)}=\frac{p(x)}{q(x)}$
Degree of $p(x)=\mathbf{2}$
Degree of $q(x)=3$
$2 < 3$
$\therefore$ It is not improper fraction.
Option (c) $\frac{x}{x^2+3 x+1}=\frac{p(x)}{q(x)}$
Degree of $p(x)=1$
Degree of $q(x)=2$
$1 < 2$
$\therefore$ It is not improper fraction.
Option (d) $\frac{p(x)}{q(x)}=\frac{x^2+1}{x^2-1}$
Degree of $p(x)=2=$ degree of $q(x)$
$\therefore \mathrm{It}$ is improper fraction.