Correct. Note that 1 and $\displaystyle-{1}\usepackage{cmbright}\Huge\$ are excluded values. Because $\displaystyle{{x}}^{{2}}-{1}={\left({x}+{1}\right)}{\left({x}-{1}\right)}\usepackage{cmbright}\Huge\$ , you should multiply both sides of the equation by $\displaystyle{\left({x}+{1}\right)}{\left({x}-{1}\right)}\usepackage{cmbright}\Huge\$ to clear the denominators:

$\displaystyle\frac{{{5}{x}{\left({{x}}^{{2}}-{1}\right)}}}{{{{x}}^{{2}}-{1}}}=\frac{{{4}{\left({x}+{1}\right)}{\left({x}-{1}\right)}}}{{{x}-{1}}}-\frac{{{2}{\left({x}+{1}\right)}{\left({x}-{1}\right)}}}{{{x}+{1}}}\usepackage{cmbright}\Huge\$

$\displaystyle{5}{x}={4}{\left({x}+{1}\right)}-{2}{\left({x}-{1}\right)}\usepackage{cmbright}\Huge\$

$\displaystyle{5}{x}={4}{x}+{4}-{2}{x}+{2}\usepackage{cmbright}\Huge\$ or $\displaystyle{5}{x}={2}{x}+{6}\usepackage{cmbright}\Huge\$

$\displaystyle{3}{x}={6}\usepackage{cmbright}\Huge\$ , so $\displaystyle{x}={2}\usepackage{cmbright}\Huge\$