Find the integral of f(x) = (x^2-3)/x for the interval [1, 2].

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.