How do you integrate #(x^2+x)/((x+2)(x-1)^2)# using partial fractions?

Question

Emma Aldridge · Accepted Answer

#int (x^2+x)/((x+2)(x-1)^2) dx =2/9 ln abs(x+2) + 7/9 ln abs(x-1) - 2/(3(x-1)) + C#

We know that we're searching for a partial fraction decomposition of the following form because the denominator has already been factored for us: #(x^2+x)/((x+2)(x-1)^2)# #=A/(x+2)+B/(x-1)+C/(x-1)^2# #=(A(x-1)^2+B(x+2)(x-1)+C(x+2))/((x+2)(x-1)^2)# #=((A+B)x^2+(-2A+B+C)x+(A-2B+2C))/((x+2)(x-1)^2)# We obtain the following system of linear equations by equating coefficients: #{ (A+B=1), (-2A+B+C=1), (A-2B+2C=0) :}# When we combine the three equations, we obtain: #3C = 2# So #color(blue)(C=2/3)# After deducting the second equation from the initial one, we obtain: #3A-C = 0# Hence #color(blue)(A = 2/9)# Then from the first equation we find #color(blue)(B=7/9)# So: #(x^2+x)/((x+2)(x-1)^2)=2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2)# Hence: #int (x^2+x)/((x+2)(x-1)^2) dx# #=int 2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2) dx# #=2/9 ln abs(x+2) + 7/9 ln abs(x-1) - 2/(3(x-1)) + C#

William Abdalla · Answer

undefined To integrate $\frac{x^2 + x}{(x+2)(x-1)^2}$ using partial fractions, first express the fraction as the sum of partial fractions. You would typically start by writing it in the form:

$\frac{x^2 + x}{(x+2)(x-1)^2} = \frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$

Then, find the values of $A$, $B$, and $C$ by equating the numerators:

$x^2 + x = A(x-1)^2 + B(x+2)(x-1) + C(x+2)$

After finding the values of $A$, $B$, and $C$, integrate each term separately, as they become simpler to integrate once expressed as partial fractions.