Evaluation of Definite Integral

For x less than -1, abs(x^2-1) = x^2-1.
For x between -1 and +1, abs(x^2-1) = 1-x^2.
So split the integral into 2 parts: from -5 to -1 and -1 to +1. We get
S(x^2-1)dx from -5 to -1 = (x^3/3 - x) from -5 to -1 = -1/3 +1 + 125/3 -5= 124/3 - 4.
S(1-x^2)dx from -1 to +1 = (x - x^3/3) from -1 to +1 = 1 - 1/3 +1 -1/3 = 2 - 2/3.
Adding together we get 122/3 -2 = 116/3 = 38and 2/3.

consider y=| x^2 -1 | ,

y= x^2-1 , -5<=x< -1

y= 1-x^2 , - 1<= x <=1

so , in(| x^2 - 1 | ) ( -5 to 1)

= in( x^2-1)(-5 to -1) + in( 1-x^2)( -1 to 1 )

= [x^3/3 -x](-5 to -1) - in [x^3/3 - x](-1 to 1)

= ( -1/3 +1)- ( -125/3+5) -( -1/3 +1) + ( 1/3 -1)

= 38 (2/3)