Answer
$${\text{The integral diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\frac{1}{{{{\left( {x - 3} \right)}^2}}}} dx \cr
& {\text{The integrand is not defined for }}x = 3,{\text{ then}} \cr
& \int_0^4 {\frac{1}{{{{\left( {x - 3} \right)}^2}}}} dx = \mathop {\lim }\limits_{b \to {3^ + }} \int_0^b {\frac{1}{{{{\left( {x - 3} \right)}^2}}}} dx + \mathop {\lim }\limits_{a \to {3^ - }} \int_a^4 {\frac{1}{{{{\left( {x - 3} \right)}^2}}}} dx \cr
& {\text{Integrating}} \cr
& \int_0^4 {\frac{1}{{{{\left( {x - 3} \right)}^2}}}} dx = \mathop {\lim }\limits_{b \to {3^ + }} \left[ {\frac{1}{{x - 3}}} \right]_0^b + \mathop {\lim }\limits_{a \to {3^ - }} \left[ {\frac{1}{{x - 3}}} \right]_a^4 \cr
& \int_0^4 {\frac{1}{{{{\left( {x - 3} \right)}^2}}}} dx = \mathop {\lim }\limits_{b \to {3^ + }} \left[ {\frac{1}{{b - 3}} - \frac{1}{{0 - 3}}} \right] + \mathop {\lim }\limits_{a \to {3^ - }} \left[ {\frac{1}{{4 - 3}} - \frac{1}{{a - 3}}} \right] \cr
& \int_0^4 {\frac{1}{{{{\left( {x - 3} \right)}^2}}}} dx = \mathop {\lim }\limits_{b \to {3^ + }} \left[ {\frac{1}{{b - 3}} + \frac{1}{3}} \right] + \mathop {\lim }\limits_{a \to {3^ - }} \left[ {1 - \frac{1}{{a - 3}}} \right] \cr
& {\text{Evaluate the limit}} \cr
& \int_0^4 {\frac{1}{{{{\left( {x - 3} \right)}^2}}}} dx = \left[ {\frac{1}{{3 - 3}} + \frac{1}{3}} \right] + \mathop {\lim }\limits_{a \to {3^ - }} \left[ {1 - \frac{1}{{3 - 3}}} \right] \cr
& \int_0^4 {\frac{1}{{{{\left( {x - 3} \right)}^2}}}} dx = \infty \cr
& {\text{The integral diverges}} \cr} $$
