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

Question

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

Paisley Johnson · Accepted Answer

That integral diverges.

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.

Charles Arocha · Answer

undefined To find the integral of $ \int \frac{x}{x^2 - 9} \, dx $ from 1 to infinity, we first perform partial fraction decomposition on the integrand:

$$ \frac{x}{x^2 - 9} = \frac{x}{(x + 3)(x - 3)} = \frac{A}{x + 3} + \frac{B}{x - 3} $$

Solving for $ A $ and $ B $, we find:

$$ A = \frac{1}{6} $$
$$ B = -\frac{1}{6} $$

So, the integral becomes:

$$ \int \frac{x}{x^2 - 9} \, dx = \int \left( \frac{1}{6} \cdot \frac{1}{x + 3} - \frac{1}{6} \cdot \frac{1}{x - 3} \right) \, dx $$

Integrating term by term, we get:

$$ \int \frac{x}{x^2 - 9} \, dx = \frac{1}{6} \ln|x + 3| - \frac{1}{6} \ln|x - 3| + C $$

Now, to find the integral from 1 to infinity, we take the limit as $ b $ approaches infinity of the definite integral from 1 to $ b $:

$$ \lim_{{b \to \infty}} \left( \frac{1}{6} \ln|b + 3| - \frac{1}{6} \ln|b - 3| \right) - \left( \frac{1}{6} \ln|1 + 3| - \frac{1}{6} \ln|1 - 3| \right) $$

Simplifying, we get:

$$ \lim_{{b \to \infty}} \left( \frac{1}{6} \ln|b + 3| - \frac{1}{6} \ln|b - 3| - \frac{1}{6} \ln 4 + \frac{1}{6} \ln 2 \right) $$

Since $ \ln|b + 3| $ and $ \ln|b - 3| $ approach infinity as $ b $ approaches infinity, the limit becomes:

$$ \lim_{{b \to \infty}} \left( \frac{1}{6} \ln b - \frac{1}{6} \ln b \right) = 0 $$

Therefore, the integral from 1 to infinity of $ \frac{x}{x^2 - 9} \, dx $ is 0.