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

Oct 19, 2015

That integral diverges.

#### Explanation:

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

${\int}_{1}^{\infty} \frac{x}{{x}^{2} - 9} \mathrm{dx} = {\int}_{1}^{3} \frac{x}{{x}^{2} - 9} \mathrm{dx} + {\int}_{3}^{\infty} \frac{x}{{x}^{2} - 9} \mathrm{dx}$
The second of which is improper at both limits of integration, so we need:

${\int}_{1}^{\infty} \frac{x}{{x}^{2} - 9} \mathrm{dx} = {\int}_{1}^{3} \frac{x}{{x}^{2} - 9} \mathrm{dx} + {\int}_{3}^{c} \frac{x}{{x}^{2} - 9} \mathrm{dx} + {\int}_{c}^{\infty} \frac{x}{{x}^{2} - 9} \mathrm{dx}$
for a chosen $c > 3$.

The first of the integrals is:

${\int}_{1}^{3} \frac{x}{{x}^{2} - 9} \mathrm{dx} = {\lim}_{b \rightarrow {3}^{-}} {\int}_{1}^{b} \frac{x}{{x}^{2} - 9} \mathrm{dx}$

$= {\lim}_{b \rightarrow {3}^{-}} {\left[\frac{1}{2} \ln \left\mid {x}^{2} - 9 \right\mid\right]}_{1}^{b}$

$= {\lim}_{b \rightarrow {3}^{-}} {\left[\frac{1}{2} \ln \left\mid {x}^{2} - 9 \right\mid\right]}_{1}^{b}$

$= {\lim}_{b \rightarrow {3}^{-}} {\left[\frac{1}{2} \ln \left(9 - {x}^{2}\right)\right]}_{1}^{b}$

$= {\lim}_{b \rightarrow {3}^{-}} \left[\frac{1}{2} \ln \left(9 - {b}^{2}\right) - \frac{1}{2} \ln 8\right]$

As $x \rightarrow {3}^{-}$, we have $\left(9 - {x}^{2}\right) \rightarrow {0}^{+}$, so $\ln \left(9 - {b}^{2}\right) \rightarrow - \infty$

The integral diverges.

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