Question

What is the integral of `(x+1)/(x(x^2+x-6))`?

Answer 1

`-1/6ln|x| - 2/15 ln|x+3| + 3/10 ln|x-2| + C`

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Well, for one, we can factor the denominator:

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

`int(x + 1)/(x(x+3)(x-2))dx -= int A/(x) + B/(x+3) + C/(x-2)dx`

Now we just have linear denominators, which is a fairly straightforward decomposition.

By achieving common denominators on the righthand side, we can equate the numerator to `x+1`. Therefore, multiply the top and bottom so that all the terms have the denominator `x(x^2 + x - 6)`.

`[A(x+3)(x-2) + B(x)(x-2) + C(x)(x+3)]/cancel(x(x+3)(x-2)) = (x+1)/cancel(x(x+3)(x-2))`

Simplify the left side so that it looks like a general polynomial. First, distribute:

`Ax^2 + Ax - 6A + Bx^2 - 2Bx + Cx^2 + 3Cx = x + 1`

Group together terms.

`Ax^2 + Bx^2 + Cx^2 + Ax - 2Bx + 3Cx - 6A = x + 1`

Now make sure you get it into this correct form, `bb(ax^2 + bx + c)`:

`ul((A + B + C))x^2 + ul((A - 2B + 3C))x + ul((-6A))`

`= ul(0)x^2 + ul(1)x + ul(1)`

This means we have a system of three equations:

`A + B + C = 0`
`A - 2B + 3C = 1`
`-6A = 1`

Clearly, `color(green)(A = -1/6)`, so the rest follows. Add the negative of the second equation to the first.

`" "A + B + C = 0`
`- (A - 2B + 3C = 1)`
`"------------------------------"`
`" "" "" "3B - 2C = -1`

This gives `C = (-1 - 3B)/(-2) = 1/2 + 3/2B`, so from the first equation:

`-1/6 + B + 1/2 + 3/2B = 0`

`1/6 - 1/2 = 5/2B`

`=> color(green)(B = -2/15)`

Thus:

`color(green)(C) = 1/2 + 3/2*-2/15 = color(green)(3/10)`

You can verify `A`, `B`, and `C` by plugging them back into the system of equations. And so, we have:

`int(x + 1)/(x^3 + x^2 - 6x)dx`

`= -1/6int 1/(x)dx - 2/15 int 1/(x+3)dx + 3/10 int 1/(x-2)dx`

We know the integral of `1/u` to be `ln|u|`. Thus:

`=> color(blue)(int(x + 1)/(x^3 + x^2 - 6x)dx)`

`color(blue)(= -1/6ln|x| - 2/15 ln|x+3| + 3/10 ln|x-2| + C)`