How do you integrate #int (4x^3 - 4x^2 - 16x + 7) / ((x + 1) (x -2))# using partial fractions?

Question

Isaiah Allen · Accepted Answer

# int \ (4x^3 - 4x^2 - 16x + 7) / ((x+1)(x-2)) \ dx = 2x^2 -5 ln |x+1| - 3ln|x-2| + C #

We seek: # I = int \ (4x^3 - 4x^2 - 16x + 7) / ((x+1)(x-2)) \ dx # As we have the algebraic equivalent of a "top-heavy" fraction—that is, the order of the numerator's polynomial is higher than the denominator's—we must first perform algebraic long division. # {: ( , , ul(+4x), ul(" "), ul(" "), ul(" "), ), ( x^2-x-2, ")", +4x^3, -4x^2, -16x, + 7, ), ( , , ul(+4x^3), ul(-4x^2), ul(-8x), ul(" "), -), ( , , , , -8x, +7, ) :} # Thus, we discover: # (4x^3 - 4x^2 - 16x + 7) / (x^2-x-2) -= 4x + (-8x+7)/(x^2-x-2) # The second term can now be broken down into partial fractions: # (-8x+7)/(x^2-x-2) -= (-8x+7)/((x+1)(x-2)) # # " " = A/(x+1) + B/(x-2) # # " " = (A(x-2) + B(x+1)) / ((x+1)(x-2)) # Getting to the answer: # -8x+7 = A(x-2) + B(x+1) # 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 = -1 => 8+7=A(-3) => A = -5# Put # x = +2 => -16+7 = 3B => B = -3# Thus, we can now write: # I = int \ 4x - 5/(x+1) - 3/(x-2) \ dx # that now includes the standard integral, which we integrate to obtain: # I = 2x^2 -5 ln |x+1| - 3ln|x-2| + C #