Question

How do you integrate `f(x)=(x^2+1)/((x^2+6)(x-4)(x-9))` using partial fractions?

Answer 1

Alright, here goes... I got:

`int (x^2+1)/((x^2+6)(x-4)(x-9))dx`

`= -65/3828 ln|x^2 + 6| - 25/(319sqrt6) arctan(x/sqrt6) - 17/110 ln|x - 4| + 82/435 ln|x - 9| + "const"`

---

DISCLAIMER: OBSCENELY LONG ANSWER!

`int (x^2 + 1)/((x^2 + 6)(x - 4)(x - 9))dx`

`= int (Ax + B)/(x^2 + 6) + C/(x - 4) + D/(x - 9)dx`

GETTING THE SYSTEM OF EQUATIONS

Now, let's focus on the integrand... get common denominators, and then multiply through by that denominator, `(x^2 + 6)(x - 4)(x - 9)`, to get:

`0x^3 + x^2 + 0x + 1`

`= (Ax + B)(x - 4)(x - 9) + C(x^2 + 6)(x - 9) + D(x^2 + 6)(x - 4)`

Now consider only the right-hand side, and distribute terms...

`= (Ax + B)(x^2 - 13x + 36) + C(x^3 - 9x^2 + 6x - 54) + D(x^3 - 4x^2 + 6x - 24)`

Keep going...

`= Ax^3 - 13Ax^2 + 36Ax + Bx^2 - 13Bx + 36B + Cx^3 - 9Cx^2 + 6Cx - 54C + Dx^3 - 4Dx^2 + 6Dx - 24D`

Now, group terms by exponent, so that you get `ax^3 + bx^2 + cx + d`, making sure that you write it in terms of addition:

`= color(green)(Ax^3 + Cx^3 + Dx^3) color(red)(- 13Ax^2 + Bx^2- 9Cx^2 - 4Dx^2) color(orange)(+ 36Ax - 13Bx + 6Cx + 6Dx) + color(darkblue)(36B - 54C - 24D)`

Now, factor it into the proper coefficients to get:

`0x^3 + x^2 + 0x + 1 = (color(green)(A + C + D))x^3 + (color(red)(- 13A + B- 9C - 4D))x^2 + (color(orange)(36A - 13B + 6C + 6D))x + (color(darkblue)(36B - 54C - 24D))`

From this, we get four equations:

`" "1A + 0B + 1C + 1D = 0`
`-13A + 1B - 9C - 4D = 1`
`" "36A - 13B + 6C + 6D = 0`
`" "0A + 36B - 54C - 24D = 1`

SOLVING THE SYSTEM OF EQUATIONS

It may be easier to rewrite this into an augmented matrix:

`[(1,0,1,1, |, 0),(-13,1,-9,-4, |, 1),(36,-13,6,6, |, 0),(0,36,-54,-24, |, 1)]`

If we reduce this down using elementary row operations, we can then more easily obtain each constant `A`, `B`, `C`, `D`.

Denote `cR_i + R_j` being to multiply row `i` by a constant `c` and adding the result to row `j`. Then:

`stackrel(-36R_1 + R_3" ")(->) [(1,0,1,1, |, 0),(-13,1,-9,-4, |, 1),(0,-13,-30,-30, |, 0),(0,36,-54,-24, |, 1)]`

`stackrel(13R_1 + R_2" ")(->) [(1,0,1,1, |, 0),(0,1,4,9, |, 1),(0,-13,-30,-30, |, 0),(0,36,-54,-24, |, 1)]`

`stackrel(13R_2 + R_3" ")(->) [(1,0,1,1, |, 0),(0,1,4,9, |, 1),(0,0,22,87, |, 13),(0,36,-54,-24, |, 1)]`

`stackrel(-36R_2 + R_4" ")(->) [(1,0,1,1, |, 0),(0,1,4,9, |, 1),(0,0,22,87, |, 13),(0,0,-198,-348, |, -35)]`

I got tired of reducing it, so we currently have:

`A + C + D = 0`
`B + 4C + 9D = 1`
`22C + 87D = 13`
`198C + 348D = 35`

Solving more explicitly, take the third equation:

`C = 13/22 - 87/22D`

Plug into the fourth:

`198(13/22 - 87/22D) + 348D = 35`

`117 - 783D + 348D = 35`

`82 = 435D`

`color(green)(D = 82/435)`

Yikes. Now, find `C`:

`22C + 87(82/435) = 13`

`22C + 82/5 = 13`

`color(green)(C) = (65 - 82)/5/22`

`= color(green)(-17/110)`

Now plug into the first equation to get `A`:

`color(green)(A) = -C - D = 17/110 - 82/435`

`= 7395/47850 - 9020/47850`

`= -1625/47850`

`= color(green)(-65/1914)`

Lastly, find `B`:

`B + 4(-17/110) + 9(82/435) = 1`

`color(green)(B) = 1 + 68/110 - 738/435`

`= 47850/47850 + (29580 - 81180)/(47850)`

`= -3750/47850`

`= color(green)(-25/319)`

SETTING UP THE RESULTANT INTEGRAL

So far, we have found:

`(A,B,C,D) = (-65/1914, -25/319, -17/110, 82/435)`

Therefore, the resultant integral is:

`int (x^2 + 1)/((x^2 + 6)(x - 4)(x - 9))dx`

`= int (-(65)/1914x - 25/319)/(x^2 + 6) -17/110 1/(x - 4) + 82/435 1/(x - 9)dx`

`= -int (65/1914x + 25/319)/(x^2 + 6)dx - 17/110 int 1/(x - 4)dx + 82/435 int 1/(x - 9)dx`

SOLVING EACH PART OF THE BEGINNING INTEGRAL

Call these integrals `I_1`, `I_2`, and `I_3`, i.e. let:

`int (x^2 + 1)/((x^2 + 6)(x - 4)(x - 9))dx = -I_1 - I_2 + I_3 + "const"`

INTEGRALS 2 AND 3

The last two are straightforward:

`I_2 = 17/110 ln|x - 4|`

`I_3 = 82/435 ln|x - 9|`

INTEGRAL 1

Now let's figure out the first one... Getting common denominators and rewriting it, we get:

`I_1 = int (65/1914x + 25/319)/(x^2 + 6)dx`

`= int (5(13x + 30))/(1914(x^2 + 6))dx`

`= 5/1914 int (13x + 30)/(x^2 + 6)dx`

`= 5/1914 int (12x)/(x^2 + 6)dx + 5/1914 int (x + 30)/(x^2 + 6)dx`

`= 5/1914 int (12x)/(x^2 + 6)dx + 5/3828 int (2x + 60)/(x^2 + 6)dx`

`= 5/1914 int (12x)/(x^2 + 6)dx + 5/3828 int (2x)/(x^2 + 6)dx + 25/319 int 1/(x^2 + 6)dx`

The first two parts of `I_1` here are via `u`-substitution (`u = x^2 + 6`, `du = 2xdx`):

`5/1914 int (12x)/(x^2 + 6)dx = 15/957 ln|x^2 + 6|`

`5/3828 int (2x)/(x^2 + 6)dx = 5/3828 ln |x^2 + 6|`

The third part of `I_1` will be an `arctan` solution of some sort, as `d/(du)[arctanu] = 1/(u^2 + 1)`.

`25/319 int 1/(x^2 + 6) = 25/319 int 1/6 1/((x//sqrt6)^2 + 1)`

`=> 25/319 int 1/(x^2 + 6)dx = 25/(319sqrt6)arctan(x/sqrt6)`

So, the integral `I_1` is:

`I_1 = int (65/1914x + 25/319)/(x^2 + 6)dx`

`= 15/957 ln|x^2 + 6| + 5/3828 ln |x^2 + 6| + 25/(319sqrt6) arctan(x/sqrt6)`

`= 65/3828 ln|x^2 + 6| + 25/(319sqrt6) arctan(x/sqrt6)`

OVERALL INTEGRAL

We finally put this all together:

`color(blue)(int (x^2 + 1)/((x^2 + 6)(x - 4)(x - 9))dx)`

`= -int (65/1914x + 25/319)/(x^2 + 6)dx - 17/110 int 1/(x - 4)dx + 82/435 int 1/(x - 9)dx`

`= -I_1 - I_2 + I_3 + "const"`

`= color(blue)(overbrace(-65/3828 ln|x^2 + 6| - 25/(319sqrt6) arctan(x/sqrt6))^(-I_1) - overbrace(17/110 ln|x - 4|)^(I_2) + overbrace(82/435 ln|x - 9|)^(I_3) + "const")`

Well, it worked! Wolfram Alpha agrees.