f(x) = (x^2-3)/x

We need to find the integral from x=1 to x= 2

First let us simplify the function.

==> f(x)= x^2/x - 3/x

==> f(x) = x - 3/x

Now we will calculate the integral.

==>F(x)= Int f(x) dx = Int x - 3/x dx

= x^2/2 - 3ln x + C

==> F(x) = (1/2)x^2 - 3*ln x

==> F(2) = (1/2) * 4 - 3ln 2 = 2-3ln2

==> F(1) = (1/2)*1 - 3ln 2= 1/2

==> The definite integral is :

F(2) - F(1) = 2-3ln2 - 1/2 = 3/2 - 3ln2.

**==> The integral for f(x) for the interval [1, 2] is (3/2) - 3*ln2.**

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

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

Int (x^2-3)dx/x = x^2/2 - 3ln x

We'll apply Leibniz Newton:

Int (x^2-3)dx/x = F(2) - F(1)

F(2) - F(1) = 4/2 - 3ln2 - 1/2 + 3ln 1

But ln 1 = 0

F(2) - F(1) = 2 - ln 8 - 1/2

F(2) - F(1) = 3/2 - ln 8

**The integral of the function (x^2-3)/x is Int (x^2-3)dx/x = 3/2 - ln 8.**