How do you integrate $\int \frac{1 - x^{2}}{(x - 9) (x - 5) (x + 12)}$ $int (1-x^2)/((x-9)(x-5)(x+12))$ using partial fractions?

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How do you integrate $\int \frac{1 - x^{2}}{(x - 9) (x - 5) (x + 12)}$ $int (1-x^2)/((x-9)(x-5)(x+12))$ using partial fractions?

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Answer 1 · 2016-09-12T18:57:12

$\int \frac{1 - x^{2}}{(x - 9) (x - 5) (x + 12)} d x$ $int (1-x^2)/((x-9)(x-5)(x+12)) dx$

$= - \frac{20}{21} ln | x - 9 | + \frac{6}{17} ln | x - 5 | - \frac{143}{357} ln | x + 12 | + constant$ $=-20/21 ln abs(x-9) + 6/17 ln abs(x-5) - 143/357 ln abs(x+12) + "constant"$

Explanation:

Let's look at the general problem:

$\int \frac{a x^{2} + b x + c}{(x - d) (x - e) (x - f)} d x$ $int (ax^2+bx+c)/((x-d)(x-e)(x-f)) dx$

where $d, e, f$ $d, e, f$ are distinct.

We can split the integrand into partial fractions like this:

$\frac{a x^{2} + b x + c}{(x - d) (x - e) (x - f)} = \frac{A}{x - d} + \frac{B}{x - e} + \frac{C}{x - f}$ $(ax^2+bx+c)/((x-d)(x-e)(x-f))=A/(x-d)+B/(x-e)+C/(x-f)$

where $A, B, C$ $A, B, C$ can be determined using Heaviside's cover up method:

$A = \frac{a d^{2} + b d + c}{(d - e) (d - f)}$ $A = (acolor(blue)(d)^2+bcolor(blue)(d)+c)/((color(blue)(d)-e)(color(blue)(d)-f))$

$B = \frac{a e^{2} + b e + c}{(e - d) (e - f)}$ $B = (acolor(blue)(e)^2+bcolor(blue)(e)+c)/((color(blue)(e)-d)(color(blue)(e)-f))$

$C = \frac{a f^{2} + b f + c}{(f - d) (f - e)}$ $C = (acolor(blue)(f)^2+bcolor(blue)(f)+c)/((color(blue)(f)-d)(color(blue)(f)-e))$

Then:

$\int \frac{a x^{2} + b x + c}{(x - d) (x - e) (x - f)} d x$ $int (ax^2+bx+c)/((x-d)(x-e)(x-f)) dx$

$= \int (\frac{A}{x - d} + \frac{B}{x - e} + \frac{C}{x - f}) d x$ $=int (A/(x-d)+B/(x-e)+C/(x-f)) dx$

$= A ln | x - d | + B ln | x - e | + C ln | x - f | + constant$ $=A ln abs(x-d) + B ln abs(x-e) + C ln abs(x-f) + "constant"$

In our example:

$a = - 1$ $a=-1$ , $b = 0$ $b = 0$ , $c = 1$ $c = 1$ , $d = 9$ $d = 9$ , $e = 5$ $e = 5$ , $f = - 12$ $f = -12$

So:

$A = \frac{- 80}{(4) (21)} = - \frac{20}{21}$ $A=(-80)/((4)(21)) = -20/21$

$B = \frac{- 24}{(- 4) (17)} = \frac{6}{17}$ $B=(-24)/((-4)(17)) = 6/17$

$C = \frac{- 143}{(- 21) (- 17)} = - \frac{143}{357}$ $C=(-143)/((-21)(-17)) = -143/357$

So

$\int \frac{1 - x^{2}}{(x - 9) (x - 5) (x + 12)} d x$ $int (1-x^2)/((x-9)(x-5)(x+12)) dx$

$= - \frac{20}{21} ln | x - 9 | + \frac{6}{17} ln | x - 5 | - \frac{143}{357} ln | x + 12 | + constant$ $=-20/21 ln abs(x-9) + 6/17 ln abs(x-5) - 143/357 ln abs(x+12) + "constant"$

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How do you integrate $\int \frac{1 - x^{2}}{(x - 9) (x - 5) (x + 12)}$ $int (1-x^2)/((x-9)(x-5)(x+12))$ using partial fractions?

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Explanation:

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