What is the area of the region below the graph of f and the lines x=1, x=2?f(x)=1/(x^2 + 5x + 6)

To compute the area of the given region, we'll have to
determine the definite integral of f(x).

We'll re-write the
denominator of f(x) as a product of linear factors.

We know
that: ax^2 + bx + c = a(x-x1)(x-x2), where x1 and x2 are the roots of the
quadratic.

We notice that x1 = 2 and x2 =
3.

f(x) =
1/(x+2)(x+3)

Int f(x)dx = Int
dx/(x+2)(x+3)

We'll apply Leibniz-Newton to calculate the
definite integral.

First, we'll calculate the indefinite
integral of f(x). To determine the indefinite integral, we'll write the function as a
sum or difference of partial fractions.

1/(x+2)(x+3) =
A/(x+2) + B/(x+3)

1 = A(x+3) +
B(x+2)

We'll remove the
brackets:

1 = Ax + 3A + Bx +
2B

We'll combine like terms:

1
= x(A+B) + 3A + 2B

A + B = 0

A
= -B

-3B + 2B = 1

-B =
1

B = -1

A =
1

1/(x+2)(x+3) = 1/(x+2) -
1/(x+3)

Int dx/(x+2)(x+3) = Int dx/(x+2)  - Int
dx/(x+3)

Int dx/(x+2)(x+3) = ln |x+2| - ln|x+3| +
C

Int dx/(x+2)(x+3) = ln
|(x+2)/(x+3)|

But, Int dx/(x+2)(x+3) = F(2) -
F(1)

F(2) = ln
|(2+2)/(2+3)|

F(2) = ln
4/5

F(1) = ln
|(1+2)/(1+3)|

F(1) = ln
3/4

F(2) - F(1) = ln 4/5 - ln
3/4

F(2) - F(1) = ln
(4/5)*(4/3)

F(2) - F(1) = ln
16/15

The area of the region located under
the graph of f(x), the lines x  =1 and x = 2 and x axis is:
A = ln 16/15 square units.