# How do you test the improper integral int (x-1)^-2+(x-3)^-2 dx from [1,3] and evaluate if possible?

Jun 8, 2018

The improper integral diverges

#### Explanation:

Compute the indefinite integral

$\int \left(\frac{1}{x - 1} ^ 2 + \frac{1}{x - 3} ^ 2\right) \mathrm{dx} = - \frac{1}{x - 1} - \frac{1}{x - 3} + C$

Compute the boundaries

${\lim}_{x \to {1}^{+}} \left(- \frac{1}{x - 1} - \frac{1}{x - 3}\right) = {\lim}_{x \to {1}^{+}} \left(- \frac{1}{x - 1}\right) - {\lim}_{x \to {1}^{+}} \left(- \frac{1}{x - 3}\right)$

$= - \infty + \frac{1}{2}$

$= - \infty$

${\lim}_{x \to {3}^{-}} \left(- \frac{1}{x - 1} - \frac{1}{x - 3}\right) = {\lim}_{x \to {3}^{-}} \left(- \frac{1}{x - 1}\right) - {\lim}_{x \to {3}^{-}} \left(- \frac{1}{x - 3}\right)$

$= - \frac{1}{2} - \left(- \infty\right)$

$= + \infty$

Finally,

${\int}_{1}^{3} \left(\frac{1}{x - 1} ^ 2 + \frac{1}{x - 3} ^ 2\right) \mathrm{dx} = \text{ diverges}$