integral from 0 to 1 of

x-4/(x^2-5x+6)

I do believe we can integrate with partial fractions, and I should factor it in order to create a linear case 1 thingy

so then, it will be:
x-4/(x-2)(x-3)

After that, I will find the value of a/(x-2) + b/(x-3).

But I suck at matrices but I am practicing. I just want to make video games

To integrate the given expression, we can indeed use partial fractions. Here's how you can approach it step by step:

Step 1: Factorize the denominator of the expression:

x^2 - 5x + 6 = (x - 2)(x - 3)

Step 2: Express the integrand as a sum of two fractions:

(x - 4)/[(x - 2)(x - 3)] = A/(x - 2) + B/(x - 3)

Step 3: Find the values of A and B by equating the numerators:

(x - 4) = A(x - 3) + B(x - 2)

Expanding and simplifying the equation, we get:

x - 4 = (A + B)x - (3A + 2B)

Equating the coefficients of x and the constants on both sides, we have:

1 = A + B (for the coefficient of x)
-4 = -3A - 2B (for the constant terms)

Step 4: Solve the equations to find A and B. There are multiple methods to solve this, such as substitution, elimination, or using matrices.

Using the substitution method:

From the first equation, we have A = 1 - B. Substituting this into the second equation, we get:

-4 = -3(1 - B) - 2B
-4 = -3 + 3B - 2B
-4 = -3 + B
B = -4 + 3
B = -1

Substituting the value of B into the first equation, we find:

1 = A + (-1)
1 = A - 1
A = 1 + 1
A = 2

Therefore, A = 2 and B = -1.

Step 5: Now that we have A and B, we can rewrite the expression as:

(x - 4)/[(x - 2)(x - 3)] = 2/(x - 2) - 1/(x - 3)

Step 6: Integrate each fraction separately:

∫[(x - 4)/[(x - 2)(x - 3)]]dx = ∫[2/(x - 2)]dx - ∫[1/(x - 3)]dx

Using the integral rules, we have:

= 2∫[1/(x - 2)]dx - ∫[1/(x - 3)]dx

= 2ln|x - 2| - ln|x - 3| + C

Therefore, the integral from 0 to 1 of (x - 4)/[(x^2 - 5x + 6)] is:

2ln|1 - 2| - ln|1 - 3| = 2ln|-1| - ln|-2|

Remember to evaluate the logarithmic functions with negative arguments as absolute values:

= 2ln(1) - ln(2)

Simplifying further:

= 2(0) - ln(2)

= -ln(2)

So, the final answer to the integral is -ln(2).