How do I evaluate this integral?

Question

#int_0^1(x^3-4x-9)/(x^2-x-6)dx#

Scarlett Allison · Accepted Answer

We seek: # I = int_0^1 \ (x^3-4x-9)/(x^2-x-6) \ dx#  As the degree of the numerator is higher than that of the denominator, then we can use algebraic long division to reduce the ordre of the fraction.   We find that: # (x^3-4x-9) divide (x^2-x-6) = (x+1) # with a remainder #(3x-3)# Thus: # I = int_0^1 \ (x+1) + (3x-3)/(x^2-x-6) \ dx#  We now decompose the second component into partial fractions: # (3x-3)/(x^2-x-6) -= (3x-3)/((x+2)(x-3)) #
# "               " = A/(x+2) + B/(x-3) #
# "               " = ( A(x-3) + B(x+2)) /((x+2)(x-3))  # Leading to: # (3x-3) -= A(x-3) + B(x+2)# Where #A,B# are constants that are to be determined.  We can find them by substitutions (In practice we do this via the "cover up" method: Put # x = -2 => -6-3 = A(-2-3) => A = 9/5 #
Put # x =  +3 => 9-3=B(3+2) => B = 6/5 # Using this we can write the integral as: # I = int_0^1 \ (x+1) + (9/5)/(x+2) + (6/5)/(x-3) \ dx#  Which we can now readily integrate to give: # I = [1/2x^2+x+9/5ln|x+2| + 6/5ln|x-3|]_0^1 #
# \ \  = (1/2+1+9/5ln3+6/5ln2) - (0+0+9/5ln2+6/5ln3) # # \ \  = 3/2 + 9/5ln3+6/5ln2 - 9/5ln2-6/5ln3 # # \ \  = 3/2 + (9/5-6/5)ln3 + (6/5-9/5)ln2 # # \ \  = 3/2 + 3/5ln3 -3/5ln2 # # \ \  = 3/2 + 3/5(ln3 -ln2) # # \ \  = 3/2 + 3/5ln(3/2) #