Answer
$$ - \frac{1}{x} - 5\ln \left| {\frac{{1 - 5x}}{x}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{x^2}\left( {1 - 5x} \right)}}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& = \int {\frac{1}{{{x^2}\left( {1 + \left( { - 5} \right)x} \right)}}} dx \cr
& {\text{The integrand has a expression in the form }}a + bu{} \cr
& {\text{Use formula 66}} \cr
& \left( {66} \right):\,\,\,\,\,\int {\frac{{du}}{{{u^2}\left( {a + bu} \right)}}} = - \frac{1}{{au}} + \frac{b}{{{a^2}}}\ln \left| {\frac{{a + bu}}{u}} \right| + C \cr
& {\text{let }}u = x,\,\,\,a = 1{\text{ and }}b = - 5 \cr
& \int {\frac{1}{{{x^2}\left( {1 + \left( { - 5} \right)x} \right)}}} dx = - \frac{1}{{\left( 1 \right)x}} + \frac{{ - 5}}{{{{\left( 1 \right)}^2}}}\ln \left| {\frac{{1 + \left( { - 5} \right)x}}{x}} \right| + C \cr
& {\text{simplifying}} \cr
& \int {\frac{1}{{{x^2}\left( {1 - 5x} \right)}}} dx = - \frac{1}{x} - 5\ln \left| {\frac{{1 - 5x}}{x}} \right| + C \cr} $$
