How do you integrate ∫x2−2x+3(x+3)(x−1)(x+3)dx using partial fractions?

1 Answer
Leland Adriano Alejandro
Jun 5, 2016

∫(x2−2x+3)⋅dx(x+3)(x−1)(x+3)=

92x+3+78⋅ln(x+3)+18⋅ln(x−1)+C

Explanation:

x2−2x+3(x+3)(x−1)(x+3)=A(x+3)2+Bx+3+Cx−1

Use LCD =(x+3)2(x−1), this is the denominator

x2−2x+3(x+3)2(x−1)=A(x−1)+B(x+3)(x−1)+C(x+3)2(x+3)2(x−1)

x2−2x+3(x+3)2(x−1)=Ax−A+Bx2+2Bx−3B+Cx2+6Cx+9C(x+3)2(x−1)

the equations are

B+C=1
A+2B+6C=−2
−A−3B+9C=3

Simultaneous solution results to

A=−92
B=78
C=18

x2−2x+3(x+3)(x−1)(x+3)=A(x+3)2+Bx+3+Cx−1

x2−2x+3(x+3)(x−1)(x+3)=−92(x+3)2+78x+3+18x−1

Let us now integrate

∫(x2−2x+3)dx(x+3)(x−1)(x+3)=

∫−92dx(x+3)2+∫78dxx+3+∫18dxx−1

∫(x2−2x+3)⋅dx(x+3)(x−1)(x+3)=

92x+3+78⋅ln(x+3)+18⋅ln(x−1)+C

God bless...I hope the explanation is useful.

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