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

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Answer 1 · 2016-12-16T19:30:36

$-1/4 ln(x+1) -ln (x-2) +5/4 ln (x-3) +c$

Before integration, partial fractions can be done as explained below
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The partial fractions would thus be

$-1/(4(x+1)) -1/(x-2)+5/(4(x-3)$

Integration is now simple $-1/4 int dx/(x+1) -int dx/(x-2) +5/4 int dx/(x-3)$

= $-1/4 ln(x+1) -ln (x-2) +5/4 ln (x-3) +c$

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