The number of integral values of x is $5x-1<(x+1{)}^2<7x-3,$ is

We have $(\mathrm{x}+1{)}^2>5\mathrm{x}-1$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}^2-3\mathrm{x}+2>0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{x}-1)(\mathrm{x}-2)>0\end{array}$

$\Rightarrow \mathrm{x}<1\text{ or }\mathrm{x}>2 \left(\mathrm{i}\right)$

And $(\mathrm{x}+1{)}^2<7\mathrm{x}-3$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}^2-5\mathrm{x}+4<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{x}-1)(\mathrm{x}-4)<0\end{array}$

$\Rightarrow 1<\mathrm{x}<4 \left(\mathrm{ii}\right)$

From (i) and (ii), common values are, $\mathrm{x}\in (2,4)$

Therefore, there is only one integral value of x which is 3.