# 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.

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f(x) = (2x-5) /(x^2 -5x + 6)

F(x)= intg f(x) = intg (2x-5)/(x^2 - 5x + 6)

Let u = x^2 - 5x + 6

==> du = (2x - 5) dx

F(x) = intg du/u

= ln u

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

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

F(0) = ln (0-0+6) = ln 6

Then the definite intergral = ln 2 - ln 6

= ln 2/6

= ln 1/3

= -ln3

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**

To evaluate the indefinit Int (2x-5)/(x^2-5x-6) dx .(ii) find the definite integral fro x= 0 to x=1.

Solution:

method (i) :

Pbviously numerator is diffrential coefficient of dinominator or

(x^2-5x-6)' = 2x-5.

So Int d(x^2-5x-6)/ (x^2-5x-6) = log(x^2-5x -6) + constant.

2nd method:

The denominator x^2-5x-6 = (x-6)(x+1)

So we split (2x-5)/(x^2-5x-6) into A/(x-6) +B/(x+1).

So (2x-5) = A(x+1) +B(x-6)....(1) Put x = 6, then

2*6-5 = A(6+1) + 0 . So 7 = 7A, So A = 1.

Put x = -1 in (1): 2*-1-5 = 0 + B(-1-6) . -7 = -7B. S B = a.

So 2x-5 = 1/(x-6) +1/(x+1)

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

Int (2x-5)/(x^2-5x-6) = log(x-6)+log(x+1) + Constant = log(x^2-5x-6) + C = F(x) say

To find the definite integral from x= 0 to x=1

F(1) - F(0) = log (1-5-6) - log (0-0-6) = ln(-10)/(-6) = ln(5/3) = 0.510825623.