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

#x^2/((x-3)^2(x+4))=A/(x+4)+B/(x-3)+C/(x-3)^2#
#x^2=A(x-3)^2+B(x-3)(x+4)+C(x+4)#

Substituting #x=3#:
#9=7C# so #C=9/7#

Substituting #x=-4#:
#16=49A# so #A=16/49#

Looking at the coefficients of #x^2# on both sides, we see that #x^2=Ax^2+Bx^2# or that #A+B=1#.
#B=1-16/49=33/49#

Therefore, the partial fraction decomposition of #x^2/((x-3)^2(x+4))# is #(16/49)/(x+4)+(33/49)/(x-3)+(9/7)/(x-3)^2#.

#intx^2/((x-3)^2(x+4))dx=int((16/49)/(x+4)+(33/49)/(x-3)+(9/7)/(x-3)^2)dx#

#int((16/49)/(x+4)+(33/49)/(x-3)+(9/7)/(x-3)^2)dx=16/49ln|x+4|+33/49ln|x-3|-9/(7(x-3))+C#

Apply partial fraction decomposition:

#x^2/((x-3)^2(x+4)) = A/(x-3)+B/(x-3)^2 +C/(x+4)#

#x^2/((x-3)^2(x+4)) = (A(x-3)(x+4)+B(x+4)+C(x-3)^2)/((x-3)^2(x+4))#

#x^2 = A(x^2+x-12)+Bx+4B+C(x^2-6x+9))#

#x^2 = Ax^2+ Ax-12A +Bx+4B +Cx^2 -6Cx+9C#

#x^2 = (A+C)x^2 + (A+B-6C)x - (12A-4B-9C)#

#{(A+C=1),(A+B-6C=0),(12A-4B-9C=0):}#

#{(A=1-C),(B-7C=-1),(4B+21C=12):}#

#{(A=1-C),(-4B+28C=4),(4B+21C=12):}#

#{(A=33/49),(B=9/7),(C=16/49):}#

Then:

#int (x^2dx)/((x-3)^2(x+4)) =33/49int dx /(x-3)+9/7int dx/(x-3)^2 +16/49int dx/(x+4)#

#int (x^2dx)/((x-3)^2(x+4)) =33/49ln abs(x-3)-9/(7(x-3)) +16/49ln abs(x+4)+C#