How do you use partial fraction decomposition to decompose the fraction to integrate #(x^2+x)/((x+2)(x-1)^2)#?

Question

Paisley Johnson · Accepted Answer

#(x^2+x)/((x+2)(x-1)^2) = (2/9)/(x+2) + (7/9)/(x-1) + (2/3)/(x-1)^2 #:

#(x^2+x)/((x+2)(x-1)^2)# We want #A/(x+2) + B/(x-1) + C/(x-1)^2 = (x^2+x)/((x+2)(x-1)^2) #: Clear the denominator to get: #A(x^2-2x+1)+B(x^2+x-2)+C(x+2) = x^2+x# So we need to solve the system: #A+B=1# #-2A+B+C=1# #A-2B+2C=0# Solve to get: #A = 2/9#, #" "B = 7/9#, and #" "C=2/3#

Emma Aldridge · Answer

undefined To decompose the fraction $\frac{x^2+x}{(x+2)(x-1)^2}$ using partial fraction decomposition, follow these steps:

1. Write the fraction in the form $\frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$.
2. Multiply both sides of the equation by the denominator $(x+2)(x-1)^2$.
3. Combine the fractions on the right side of the equation into a single fraction.
4. Equate the numerators on both sides of the equation.
5. Solve the resulting system of equations for the unknowns $A$, $B$, and $C$.
6. Once you have found the values of $A$, $B$, and $C$, substitute them back into the decomposed expression.