Solve first the indefinite integral, noting that:
#x^2-x-6 = (x-3)(x+2)#
#x^3-4x-10 = x(x^2-4) -10 = x (x-2) (x+2) -10#
So:
#int (x^3-4x-10)/(x^2-x-6)dx = int (x (x-2) (x+2) -10)/((x-3)(x+2))dx#
Simplify and use the linearity of the integral:
#int (x^3-4x-10)/(x^2-x-6) dx = int (x (x-2) )/(x-3) dx -10int dx/((x-3)(x+2))#
Solve the integrals separately. In the first substitute #(x-3) = u#:
#int (x (x-2) )/(x-3) dx = int ((u+3)(u+1))/udu#
#int (x (x-2) )/(x-3) dx = int (u^2+4u +3)/udu#
and always using linearity:
#int (x (x-2) )/(x-3) dx = int udu +4 int du +3int (du)/u#
#int (x (x-2) )/(x-3) dx = u^2/2 +4 u +3 ln abs u#
and undoing the substitution:
#int (x (x-2) )/(x+3) dx = (x-3)^2/2 +4 (x-3) +3 ln abs (x-3)+C#
Solve the second integral with partial fraction decomposition:
#1/((x-3)(x+2)) = A/(x-3)+B/(x+2)#
#1 = A(x+2) +B(x-3)#
#1 = (A+B)x + (2A -3B)#
#{(A+B=0),(2A-3B=1):}#
#{(A=1/5),(B=-1/5):}#
So:
#int dx/((x-3)(x+2)) = 1/5 int dx/(x-3) -1/5 int dx/(x+2)#
#int dx/((x-3)(x+2)) = 1/5ln abs(x-3) -1/5 ln abs(x+2)+C#
Put together the partial results:
#int (x^3-4x-10)/(x^2-x-6)dx = (x-3)^2/2 +4 (x-3) +3 ln abs (x-3)-2 ln abs(x-3) +2 ln abs(x+2)+C#
Simplifying and absorbing the constants in #C#:
#int (x^3-4x-10)/(x^2-x-6)dx = x^2/2 +x + ln abs (x-3)+2 ln abs(x+2)+C#
Now:
#int_0^1 (x^3-4x-10)/(x^2-x-6)dx = 1/2 +1 +ln 2+2ln 3 - ln3 -2ln2#
#int_0^1 (x^3-4x-10)/(x^2-x-6)dx = 3/2 - ln 2+ln 3 #
#int_0^1 (x^3-4x-10)/(x^2-x-6)dx = 3/2 + ln( 3/2) #
