Find the definite integral using part decomposition.

Question

∫(from 0 to 1) (x-4)/(x²-5x+6)

Answer 1

Using Partial Fraction Decomposition:

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

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

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

A+B = 1

-2A-3B = -4

Multiply the first equation by 2:

2A + 2B = 2

-2A - 3B = -4

So, B = 2 and A = -1.

∫_{(0 to 1)} [(x-4) / (x²-5x+6)]dx = ∫_{(0 to 1)} [-1 / (x-3) + 2 / (x-2)] dx

= [2ln l x-2 l - ln l x-3 l]_{(0 to 1)}

= (2 ln1 - ln2) - (2ln2 - ln3) = ln3 - 3ln2 = ln3 - ln8 = ln(3/8)