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

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Answer 1 · 2016-04-19T21:22:37

$-1/6lnabs(x+3)-lnabs(x-2)+7/6lnabs(x-3)+C$

First, focus on writing the fraction in terms of partial fractions:

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

Multiply both sides by $(x+3)(x-2)(x-3)$ :

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

Expand:

$2x+1=A(x^2-5x+6)+B(x^2-9)+C(x^2+x-6)$

Sort by degree:

$2x+1=x^2(A+B+C)+x(-5A+C)+(6A-9B-6C)$

Giving the system:

${(A+B+C=0),(-5A+C=2),(6A-9B-6C=1):}$

Solving which gives:

${(A=-1/6),(B=-1),(C=7/6):}$

Thus, we know that

$int(2x+1)/((x+3)(x-2)(x-3))dx=int-1/(6(x+3))-1/(x-2)+7/(6(x-3))dx$

Splitting up the integral and refactoring constants:

$=-1/6int1/(x+3)dx-int1/(x-2)dx+7/6int1/(x-3)dx$

$=-1/6lnabs(x+3)-lnabs(x-2)+7/6lnabs(x-3)+C$

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