To solve this question we can break it down in to a multiple small stages.

The first thing to tackle is the use of modulus around 5x - 6. This effectively means we have two equations to solve:
EQ1: x² ≥ 5x - 6
EQ2: -x² ≤ 5x - 6

Let's start by dealing with EQ1. We can first simplify it to x² - 5x + 6 ≥ 0. The second thing we should notice is that we have an inequality. Let's first find solutions for x² - 5x + 6 = 0. To do this we can either look for simple factorisations or use the quadratic formula. We'll use the quadratic formula since that will work in all situations:

x = (-(-5) ± √((-5)² - 4 * 6)) / 2

It's a bit messy to type out but simplifies easily to:

x = (5 ± 1) / 2

Giving us x = 2, 3. Now we should recall that we wanted to solve x² - 5x + 6 ≥ 0 not x² - 5x + 6 = 0 and so we need to check where the equation is greater than 0. The easiest way to do this is by plugging in numbers around our solutions, let's try 0 and 4:

0² - 5(0) + 6 = 6
4² - 5(4) + 6 = 2

So the equation is greater than 0 for x ≤ 2 and x ≥ 3 and these are our solutions to EQ1.

Now let's move on to EQ2. We can first simplify it to x² + 5x - 6 ≥ 0. Let's again look at finding solutions to x² + 5x - 6 = 0. Using the quadratic formula as before we find that x = 1, -6. As with EQ1 we should now look around these solutions to find where the equation is greater than 0, let's try 0:

0² + 5(0) - 6 = -6

The equation is less than 0 between -6 and 1, this means it is greater than 0 outside these solutions. So the equation is greater than 0 for x ≤ -6 and x ≥ 1.

Now we have the bounds on our two individual equations we need to combine them. For x² ≥ | 5x - 6 | to be true all of our inequalities must be satisfied at once. Let's list our inequalities in numerical order, we have:

x ≤ -6, x ≥ 1, x ≤ 2, x ≥ 3

Where we have signs pointing away from each other we don't need to change anything, but for the middle two we have signs pointing towards each other so we should combine these two inequalities to get our final solution of:

x ≤ -6, 1 ≤ x ≤ 2, x ≥ 3

We can do a few quick sanity checks to make sure we haven't made mistakes. It definitely looks like a solution to the question, we're using ≥/≤ as the question did, and we have multiple regions as we'd expect from a modular question. Further, if we had time, we could plug some numbers into the original equation to make sure they gave the expected results.

integrate x^2 + ln(x)

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