The area bounded by the curve $x=\left|y^2-1\right|$ and the line $y=x-5$ is

$y=x-5,y^2=1+x$ meet $y=-2,3$

$\therefore$ Required area $={\int }_{-2}^{-1} \left(y+5-\left(y^2-1\right)dy\right.$

$+{\int }_{-1}^1 \left(y+5-\left(1-y^2\right)\right)dy+{\int }_1^3 \left(y+5-\left(y^2-1\right)\right)dy$

$\begin{array}{l}={\int }_{-2}^{-1} \left(y+6-y^2\right)dy+{\int }_{-1}^1 \left(y+4+y^2\right)dy+{\int }_1^3 \left(y+6-y^2\right)dy\\ =\frac{13}{6}+\frac{26}{3}+\frac{22}{3}=\frac{109}{6}.\end{array}$