Answer
$$\frac{1}{{25}}\left( {\frac{4}{{4 - 5x}} + \ln \left| {4 - 5x} \right|} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{{\left( {4 - 5x} \right)}^2}}}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& = \int {\frac{{xdx}}{{{{\left( {4 + \left( { - 5} \right)x} \right)}^2}}}} \cr
& {\text{The integrand has a expression in the form }}a + bu{} \cr
& {\text{Use the formula 62}} \cr
& \left( {62} \right):\,\,\,\int {\frac{{udu}}{{{{\left( {a + bu} \right)}^2}}}} = \frac{1}{{{b^2}}}\left[ {\frac{a}{{a + bu}} + \ln \left| {a + bu} \right|} \right] + C \cr
& {\text{let }}u = x,\,\,\,a = 4{\text{ and }}b = - 5 \cr
& \int {\frac{{xdx}}{{{{\left( {4 + \left( { - 5} \right)x} \right)}^2}}}} = \frac{1}{{{{\left( { - 5} \right)}^2}}}\left[ {\frac{4}{{4 + \left( { - 5} \right)x}} + \ln \left| {4 + \left( { - 5} \right)x} \right|} \right] + C \cr
& {\text{simplifying}} \cr
& \int {\frac{x}{{{{\left( {4 - 5x} \right)}^2}}}} dx = \frac{1}{{25}}\left( {\frac{4}{{4 - 5x}} + \ln \left| {4 - 5x} \right|} \right) + C \cr} $$
