Write the partial fractions equation:
7x−2(x−3)2(x+1)=Ax+1+Bx−3+C(x−3)2
Multiply both sides by (x−3)2(x+1):
7x−2=A(x−3)2+B(x−3)(x+1)+C(x+1)
Let x=−1
7(−1)−2=A(−1−3)2+B(−1−3)(−1+1)+C(−1+1)
−9=16A
A=−916
7x−2=−916(x−3)2+B(x−3)(x+1)+C(x+1)
Let x=3
7(3)−2=−916(3−3)2+B(3−3)(3+1)+C(3+1)
#C = 19/4
7x−2=−916(x−3)2+B(x−3)(x+1)+194(x+1)
Let x=0
7(0)−2=−916(0−3)2+B(0−3)(0+1)+194(0+1)
−2=−8116−3B+194(0+1)
B=916
The partial fractions are:
7x−2(x−3)2(x+1)=−9161x+1+9161x−3+1941(x−3)2
Integrate:
∫7x−2(x−3)2(x+1)dx=−916∫1x+1dx+916∫1x−3dx+194∫1(x−3)2dx
The integrals are trivial:
∫7x−2(x−3)2(x+1)dx=−916lnt|x+1|+916ln|x−3|−1941x−3+C
