Question

How do you integrate `int9/((x^2+9)(x+3)(x-3)) dx` using partial fractions?

Answer 1

`- 1 /6 arctan(x/3) - 1/12 ln abs(x+3) + 1/12 ln abs(x-3) + c`

Explanation:

Your partial fraction decomposition can be computed as follows:

Find `A`, `B`, `C` and `D` so that

`9 / ((x^2+9)(x+3)(x-3)) = (Ax + B)/(x^2+9) + C / (x+3) + D / (x-3)`

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To solve this equation for `A`, `B`, `C` and `D`, the first thing to do would be multiplying both sides with `(x^2+9)(x+3)(x-3)`:

`9 = (Ax + B)(x+3)(x-3) + C(x^2+9)(x-3) + D(x^2+9)(x+3)`

... expand the terms...

`9 = (Ax + B) (x^2 -9) + C (x^3 - 3x^2 + 9x - 27) + D(x^3 + 3x^2 + 9x + 27)`

`9 = Ax^3 + Bx^2 -9Ax - 9B + Cx^3 - 3Cx^2 + 9Cx - 27C + Dx^3 + 3Dx^2 + 9Dx + 27D`

... sort the `color(red)(x^3)` terms, `color(blue)(x^2)` terms, `color(violet)(x)` terms and terms `color(green)("without " x)`...

`color(red)(0 * x^3) + color(blue)(0 * x^2) + color(violet)(0 *x) + color(green)(9) = color(red)(Ax^3) + color(blue)(Bx^2) color(violet)(- 9Ax) color(green)(- 9B) + color(red)(Cx^3) color(blue)(- 3Cx^2) + color(violet)(9Cx) color(green)(- 27C) + color(red)(Dx^3) + color(blue)(3Dx^2) + color(violet)(9Dx) + color(green)(27D)`

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This equation can only hold if all the `x^3` terms match, all the `x^2` terms match, all the `x` terms match and all the terms without `x` match.

Thus, it can be split up into `4` equations:

`{ ( (I) color(white)(xxx) 0 = color(white)(xx) A color(white)(xxxxxxi) + C color(white)(xxi)+ D color(white)(xxxxxxxx) color(red)(x^3) " terms" ), ( (II) color(white)(xx) 0 = color(white)(xxxxxx)B color(white)(xx)- 3C color(white)(xx) + 3D color(white)(xxxxxxiii) color(blue)(x^2) " terms" ), ( (III) color(white)(xii) 0 = -9A color(white)(xxxxxii) + 9C color(white)(xx)+ 9D color(white)(xxxxxxiii) color(violet)(x) " terms" ), ( (IV) color(white)(xx) 9 = color(white)(xxxx)-9B color(white)(x) -27C color(white)(x)+ 27D color(white)(xxx) color(green)("without " x )" terms") :}`

First of all, let's divide the equations `(III)` and `(IV)` by `9`:

`{ ( (I) color(white)(xxx) 0 = color(white)(xx) A color(white)(xxxxxxi) + C color(white)(xxi)+ D ), ( (II) color(white)(xx) 0 = color(white)(xxxxxx)B color(white)(xx)- 3C color(white)(xx) + 3D), ( (III) color(white)(xii) 0 = -A color(white)(xxxxxxi) + C color(white)(xxi)+ D), ( (IV) color(white)(xx) 1 = color(white)(xxxx)-B color(white)(xx) -3C color(white)(xx)+ 3D) :}`

`(I) - (III)` gives

`0 = 2A " "=>" " A = 0`

`(II) - (IV)` gives

`-1 = 2B " "=>" " B = - 1 / 2`

Inserting `A = 0` in `(I)` and `B = - 1 /2` in `(II)` gives:

`{ ( (I') color(white)(xxx) 0 = color(white)(xx) C + D), ( (II') color(white)(xx) 1/2 = -3C + 3D):}`

From `(I')` we know that `C = -D`. Inserting this into `(II')` gives

`1/2 = 3D + 3D " " => " " D = 1/12`

and finally, `C = - 1 / 12`.

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Now that we have found `A`, `B`, `C` and `D`, we can apply the partial fraction decomposition and solve the integral:

`int 9 / ((x^2+9)(x+3)(x-3)) "d" x`

`= int (-1/2 * 1 / (x^2 +9) - 1/12 * 1/(x+3) + 1/12 * 1 /(x-3)) "d"x`

`= -1/2 int 1 / (x^2 +9) "d"x - 1/12 int 1/(x+3) "d"x + 1/12 int 1 /(x-3) "d"x`

Now, you need to compute three easier integrals instead of one complicated one.

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Since

`(arctan x)' = 1 / (x^2+1)`,

we know that

`[1/3 arctan (x/3)] ' = 1/3 * 1 /((x/3)^2 + 1) = 1/(3 (x^2/9 + 3)) = 1 / (x^2 + 9)`.

Also, we know that

`(ln abs(x+1))' = 1 / (x+1)`

Thus, we can solve the integral as follows:

`int 9 / ((x^2+9)(x+3)(x-3)) "d" x`

`= -1/2 int 1 / (x^2 +9) "d"x - 1/12 int 1/(x+3) "d"x + 1/12 int 1 /(x-3) "d"x`

`= - 1 /2 * 1 / 3 * arctan(x/3) - 1/12 ln abs(x+3) + 1/12 ln abs(x-3) + c`

`= - 1 /6 arctan(x/3) - 1/12 ln abs(x+3) + 1/12 ln abs(x-3) + c`