How do you integrate $int (x^2-2x+3)/((x+3)(x-1)(x+3)) dx$ using partial fractions?

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Answer 1 · 2016-06-05T03:11:04

$int ((x^2-2x+3)*dx)/((x+3)(x-1)(x+3))=$

$(9/2)/(x+3)+7/8*ln(x+3)+1/8*ln(x-1)+C$

$(x^2-2x+3)/((x+3)(x-1)(x+3))=A/(x+3)^2+B/(x+3)+C/(x-1)$

Use LCD $=(x+3)^2(x-1)$ , this is the denominator

$(x^2-2x+3)/((x+3)^2(x-1))=(A(x-1)+B(x+3)(x-1)+C(x+3)^2)/((x+3)^2(x-1))$

$(x^2-2x+3)/((x+3)^2(x-1))=(Ax-A+Bx^2+2Bx-3B+Cx^2+6Cx+9C)/((x+3)^2(x-1))$

the equations are

$B+C=1$
$A+2B+6C=-2$
$-A-3B+9C=3$

Simultaneous solution results to

$A=-9/2$
$B=7/8$
$C=1/8$

$(x^2-2x+3)/((x+3)(x-1)(x+3))=A/(x+3)^2+B/(x+3)+C/(x-1)$

$(x^2-2x+3)/((x+3)(x-1)(x+3))=(-9/2)/(x+3)^2+(7/8)/(x+3)+(1/8)/(x-1)$

Let us now integrate

$int ((x^2-2x+3)dx)/((x+3)(x-1)(x+3))=$

$int (-9/2dx)/(x+3)^2+int (7/8dx)/(x+3)+int (1/8dx)/(x-1)$

$int ((x^2-2x+3)*dx)/((x+3)(x-1)(x+3))=$

$(9/2)/(x+3)+7/8*ln(x+3)+1/8*ln(x-1)+C$

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

How do you integrate int (x^2-2x+3)/((x+3)(x-1)(x+3)) dxint (x^2-2x+3)/((x+3)(x-1)(x+3)) dx using partial fractions?