How do you integrate #(x^3-2x^2-4)/(x^3-2x^2)# using partial fractions?

Question

Charles Anctil · Accepted Answer

THe answer is #=x-2/x+lnx-ln(x-2)+C# Let's start by rewriting the expression #(x^3-2x^2-4)/(x^3-2x^2)=1-4/(x^3-2x^2)=1-4/((x^2)(x-2))# So now we can apply the decomposition into partial fractions #1/((x^2)(x-2))=A/x^2+B/x+C/(x-2)# #=(A(x-2)+Bx(x-2)+Cx^2)/((x^2)(x-2))# Solving for A,B and C #1=A(x-2)+Bx(x-2)+Cx^2# let #x=2##=>##1=4C##=>##C=1/4# let #x=0##=>##1=-2A##=>##A=-1/2# Coefficients of #x^2##=>##0=B+C##=>##B=-1/4# so we have, #1/((x^2)(x-2))=-1/(2x^2)-1/(4x)+1/(4(x-2)# So #1-4/((x^2)(x-2))=1-2/(x^2)-1/(x)+1/(x-2)# #int((x^3-2x^2-4)dx)/(x^3-2x^2)=int(1+2/(x^2)+1/(x)-1/(x-2))dx# #=x-2/x+lnx-ln(x-2)+C#

Isabella Ackerman · Answer

undefined To integrate $ \frac{x^3 - 2x^2 - 4}{x^3 - 2x^2} $ using partial fractions, first factor the denominator and express the fraction in partial fraction form. Factoring the denominator gives $ x^2(x - 2) $. So, the partial fraction decomposition is:

$$ \frac{x^3 - 2x^2 - 4}{x^3 - 2x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 2} $$

Now, multiply both sides by the denominator $ x^3 - 2x^2 $ to clear the fractions and solve for $ A $, $ B $, and $ C $. After solving for the constants, integrate each term separately.