Evaluate the definite integral of f(x)=x^2/(x^2 - 6x + 10) from x=3/2 to x=2.

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

We'll calculate Integral
of f(x) = x^2/(x^2 - 6x + 10)

We notice that if we'll
complete the numrator, we'll obtain the denominator, x^2 - 6x + 10, so, we'll subtract
and add 6x and we'll add and subtract 10.

The numerator
will become:

x^2 - 6x + 10 + 6x -
10

The integral will
become:

Int (x^2- 6x + 10 + 6x - 10)dx/(x^2 - 6x +
10)

We'll re-write the integral, using the addition
property of integrals:

Int (x^2- 6x + 10)dx/(x^2 - 6x +
10)+Int (6x - 10)dx/(x^2 - 6x + 10)

We'll solve the first
integral:

Int (x^2- 6x + 10)dx/(x^2 - 6x + 10) = Int dx = x
+ C

We'll solve the second
integral:

Int (6x - 10)dx/(x^2 - 6x +
10)

We'll create the square (2x-3)^2 to the
denominator.

x^2 - 6x + 10 = (2x-3)^2 +
31

We'll note 2x-3= t.

We'll
write the numerator in t:

2x-3=
t

2x = t+3

x =
(t+3)/2

6(t+3)/2 - 10 = 3(t+3) -
10

We'll remove the
brackets:

3t+9-10

The
numerator will become:

6x - 10 =
3t-1

We'll re-write the integral in
t:

Int (6x - 10)dx/[(2x-3)^2 + 31] = Int (3t-1)dt/(t^2 +
31)

Int (3t-1)dt/(t^2 + 31) = 3Int dt/(t^2 + 31) - Int
dt/(t^2 + 31)

3Int dt/(t^2 + 31) = 6/2 ln (t^2 + 31) +
C

Int dt/(t^2 + 31) = (2/sqrt31)*arctan
(1/sqrt31)

Int f(x)dx = 1/3 + 3ln (32/31) -
(2/sqrt31)*arctan (1/sqrt31)