Question

How do you find the integral of `int x / (x^2-9) dx` from 1 to infinity?

Answer 1

That integral diverges.

Explanation:

Because the integrand is undefined at `x=3`, we have to try to evaluate by evaluating two improper integrals:

`int_1^oo x / (x^2-9) dx= int_1^3 x / (x^2-9) dx+int_3^oo x / (x^2-9) dx`
The second of which is improper at both limits of integration, so we need:

`int_1^oo x / (x^2-9) dx= int_1^3 x / (x^2-9) dx+int_3^c x / (x^2-9) dx + int_c^oo x / (x^2-9) dx`
for a chosen `c > 3`.

The first of the integrals is:

`int_1^3 x / (x^2-9) dx = lim_(brarr3^-) int_1^b x / (x^2-9) dx`

`= lim_(brarr3^-) [1/2ln abs(x^2-9)]_1^b`

`= lim_(brarr3^-) [1/2ln abs(x^2-9)]_1^b`

`= lim_(brarr3^-) [1/2ln (9-x^2)]_1^b`

`= lim_(brarr3^-) [1/2ln (9-b^2) - 1/2ln8]`

As `xrarr3^-`, we have `(9-x^2) rarr 0^+`, so `ln(9-b^2) rarr -oo`

The integral diverges.

Because one of the integrals diverges, the entire integral diverges.