How do you integrate $int (x-3x^2)/((x+9)(x-5)(x-7))$ using partial fractions?

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Answer 1 · 2017-10-24T11:33:51

$ln(((x+9)^(9/8)(x-5)^(5/2))/(x-7)^(7/4)) + C$

First you'll need to split up the integrand into partial fractions.

$(x-3x^2)/((x+9)(x-5)(x-7)) = A/(x+9)+B/(x-5)+C/(x-7)$

$x-3x^2 = A(x-5)(x-7) + B(x+9)(x-7) + C(x+9)(x-5)$

We'll substitute in the roots, $x=-9,5,7$ , to find each of the numerator constants.

$x = -9$

$(-9)-3xx(-9)^2 = A xx (-9-5) xx (-9-7)$

$-252 = A xx -14 xx -16$

$A = 9/8$

$x = 5$

$(5)-3(5)^2 = B xx (5+9) xx (5-7)$

$-70 = B xx 14 xx -2$

$B = 5/2$

$x = 7$

$(7)-3(7)^2 = C xx (7 + 9) xx (7-2)$

$-140 = C xx 16 xx 5$

$C = -7/4$

Therefore, we now have that

$(x-3x^2)/((x+9)(x-5)(x-7)) = (9/8)/(x+9)+(5/2)/(x-5)-(7/4)/(x-7)$

Integrating $a/(x+b)$ gives you $aln(x+b)$ , so

$int (9/8)/(x+9)+(5/2)/(x-5)-(7/4)/(x-7)dx =$

$9/8ln(x+9) + 5/2ln(x-5) - 7/4ln(x-7) + C$

$= ln(((x+9)^(9/8)(x-5)^(5/2))/(x-7)^(7/4)) + C$

How do you integrate int (x-3x^2)/((x+9)(x-5)(x-7)) int (x-3x^2)/((x+9)(x-5)(x-7)) using partial fractions?