Question

How do you find the indefinite integral of `int 2x ln(x^2 - 8x + 18) dx`?

Answer 1

This is going to be long. The answer is:

`= color(blue)((x^2 - 14) ln(x^2 - 8x + 18) + 16sqrt2arctan((x-4)/(sqrt2)) - x^2 - 8x + C)`

The difficult/potentially confusing things you will need to know for this:

• Integration by Parts (basic understanding)
• Long Division or Synthetic Division (concept only)
• Completing the Square
• Keeping track of parentheses
• u-substitution (basic understanding)
• Integral of `1/(1+u^2)`

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First off, you should notice that this integral involves one term you can differentiate easily (`ln(x^2 - 8x + 18)`) and one that you can integrate easily (`2x`). Let's try integration by parts. First, let's factor out that `\mathbf (2)`. We have:

`int xln(x^2 - 8x + 18)dx`

Let:
`u = ln(x^2 - 8x + 18)`
`du = (2x - 8)/(x^2 - 8x + 18)dx`
`dv = xdx`
`v = x^2/2`

`int udv = uv - intvdu`

`= x^2/2 ln(x^2 - 8x + 18) - int x^2/2 * (2x - 8)/(x^2 - 8x + 18)dx`

`= color(green)(x^2/2 ln(x^2 - 8x + 18)) - int color(green)(x^3 - 4x^2)/(x^2 - 8x + 18)dx`

Now this suggests that we have to divide out `(x^3 - 4x^2)/(x^2 - 8x + 18)` somehow. To save room, I will not be doing the long division here; instead, I will try multiplying the denominator by potential candidates.

I already tried multiplying the denominator by `x-4` and it didn't work, so I'm going to try multiplying by `\mathbf(x + 4)` to see if I can get the denominator to look like `x^3 - 4x^2` so we can divide this out.

`(x+4)(x^2 - 8x + 18)`

`= x^3 - 8x^2 + 18x + 4x^2 - 32x + 72`

`= color(green)(x^3 - 4x^2) - 14x + 72`

Okay, that's more like it. That means we can add `\mathbf(14x)` and subtract `\mathbf(72)` in the denominator to get the numerator to divide out properly. So far we have:

`x^3 - 4x^2 = (x+4)(x^2 - 8x + 18) + 14x - 72`

So, let's substitute back in to get:

`((x+4)(x^2 - 8x + 18) + 14x - 72)/(x^2 - 8x + 18)`

`((x+4)cancel((x^2 - 8x + 18)))/cancel(x^2 - 8x + 18) + (14x - 72)/(x^2 - 8x + 18)`

`= color(green)((14x - 72)/(x^2 - 8x + 18) + x + 4)`

Currently therefore, we have:

`= x^2/2 ln(x^2 - 8x + 18) - int (14x - 72)/(x^2 - 8x + 18) + x + 4dx`

We can do u-substitution here. Notice how `d/(dx)[x^2 - 8x + 18] = d/(dx)[x^2 - 8x] = 2x - 8`. Let's get that into the numerator. Since `14x` is divisible by `7`, we need to get `7(2x - 8)`, which is `14x - 56`. Thus, we need to add and subtract `16` to get the factor we want for our substitution.

`= int (14x - 56 - 16)/(x^2 - 8x + 18)dx`

`= color(green)(int (7(2x - 8))/(x^2 - 8x + 18)dx - 16int 1/(x^2 - 8x + 18)dx)`

With this first one, the integral is fairly simple. Substitute to get:

`int 7/u du`

`= color(green)(7ln(x^2 - 8x + 18))`
(as it turns out, it has no real solutions, so the absolute value bars aren't necessary.)

For the second integral, it would be really nice if it looked like `\mathbf(d/(dx)[arctanx])`... Complete the square? `(8/2)^2 = 16`, so we can subtract and then add `2`.

`16*1/(x^2 - 8x + 18)`

`= 16*1/((x - 4)^2 + 2)`

To get it to look like `\mathbf(1/("something"^2 + 1))` is the goal. Let's divide by `sqrt2` on the inside of the parentheses. Notice how we just multiplied by `"1/"("1/"(sqrt2))`, so we have to compensate by dividing by `sqrt2` on the outside of the integral.

`= 16*1/sqrt2 int 1/(((x-4)/sqrt2)^2 + 2/2)`

`= 16*1/sqrt2 int 1/(((x-4)/sqrt2)^2 + 1)`

`= color(green)(16*(arctan((x-4)/(sqrt2)))/sqrt2)`.

That's all of the hard ones! Overall, we have just done this:

`= x^2/2 ln(x^2 - 8x + 18) - int (x^3 - 4x^2)/(x^2 - 8x + 18)dx`

`= x^2/2 ln(x^2 - 8x + 18) - int (14x - 72)/(x^2 - 8x + 18) + x + 4dx`

`= x^2/2 ln(x^2 - 8x + 18) - [7ln(x^2 - 8x + 18) - (16arctan((x-4)/(sqrt2)))/sqrt2 + color(green)(x^2/2 + 4x)]`

`= x^2/2 ln(x^2 - 8x + 18) - 7ln(x^2 - 8x + 18) + 8sqrt2arctan((x-4)/(sqrt2)) - x^2/2 - 4x`

ALMOST THERE! Now we finish by re-multiplying by `\mathbf(2)`, since we divided by `2` in the first step. We get:

`= x^2 ln(x^2 - 8x + 18) - 14ln(x^2 - 8x + 18) + 16sqrt2arctan((x-4)/(sqrt2)) - x^2 - 8x`

`= color(blue)((x^2 - 14) ln(x^2 - 8x + 18) + 16sqrt2arctan((x-4)/(sqrt2)) - x^2 - 8x + C)`