Solve first the indefinite integral, noting that:
x^2-x-6 = (x-3)(x+2)x2−x−6=(x−3)(x+2)
x^3-4x-10 = x(x^2-4) -10 = x (x-2) (x+2) -10x3−4x−10=x(x2−4)−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∫x3−4x−10x2−x−6dx=∫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))∫x3−4x−10x2−x−6dx=∫x(x−2)x−3dx−10∫dx(x−3)(x+2)
Solve the integrals separately. In the first substitute (x-3) = u(x−3)=u:
int (x (x-2) )/(x-3) dx = int ((u+3)(u+1))/udu∫x(x−2)x−3dx=∫(u+3)(u+1)udu
int (x (x-2) )/(x-3) dx = int (u^2+4u +3)/udu∫x(x−2)x−3dx=∫u2+4u+3udu
and always using linearity:
int (x (x-2) )/(x-3) dx = int udu +4 int du +3int (du)/u∫x(x−2)x−3dx=∫udu+4∫du+3∫duu
int (x (x-2) )/(x-3) dx = u^2/2 +4 u +3 ln abs u∫x(x−2)x−3dx=u22+4u+3ln|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∫x(x−2)x+3dx=(x−3)22+4(x−3)+3ln|x−3|+C
Solve the second integral with partial fraction decomposition:
1/((x-3)(x+2)) = A/(x-3)+B/(x+2)1(x−3)(x+2)=Ax−3+Bx+2
1 = A(x+2) +B(x-3)1=A(x+2)+B(x−3)
1 = (A+B)x + (2A -3B)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)
