Solve the indefinite integral of f(x)=(2x-5)/(x^2-5x+6) . Then solve the definite integral of f from x=0 to x=1.

To calculate the indefinite integral of the given
function, we'll use the substitution method.

We'll
calculate Integral of f(x) = (2x-5)/(x^2-5x+6)

We notice
that if we'll differentiate the denominator, x^2-5x+6, we'll get
2x-5.

So, we'll note x^2-5x+6 =
t

(x^2-5x+6)'dx = dt

(2x-5)dx
= dt

We'll re-write the integral in the variable
t:

Int  (2x-5)dx/(x^2-5x+6) = Int dt /
t

Int dt / t = ln t + C

But
x^2-5x+6 = t.

Int  (2x-5)dx/(x^2-5x+6) =
ln(x^2-5x+6) + C, where C is a family of
constants.

 Now,
we'll evaluate the definite integral, using Leibniz-Newton
formula:

Int  (2x-5)dx/(x^2-5x+6) = F(1) -
F(0)

F(1) = ln(1^2-5*1+6)

F(1)
= ln (1-5+6)

F(1) = ln 2

F(0)
= ln(0^2-5*0+6)

F(0) = ln
6

Int  (2x-5)dx/(x^2-5x+6) = ln 2 - ln
6

We'll apply the quotient rule to the difference of
logarithms:

 Int  (2x-5)dx/(x^2-5x+6) = ln
(2/6)

Int (2x-5)dx/(x^2-5x+6)=ln (1/3)=ln 1 -
ln 3=0-ln 3=- ln 3