Answer
$$\frac{4}{{25\left( {4 + 5x} \right)}} + \frac{1}{{25}}\ln \left| {4 + 5x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{{\left( {4 + 5x} \right)}^2}}}} dx \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\frac{u}{{{{\left( {a + bu} \right)}^2}}}} du = \frac{1}{{{b^2}}}\left( {\frac{a}{{a + bu}} + \ln \left| {a + bu} \right|} \right) + C,{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Comparing}} \cr
& \int {\frac{u}{{{{\left( {a + bu} \right)}^2}}}} du = \int {\frac{x}{{{{\left( {4 + 5x} \right)}^2}}}} dx \to a = 4,{\text{ }}b = 5,{\text{ }}u = x \cr
& {\text{Then substituting the values of }}a,b{\text{ and }}x{\text{ into }}\left( {\bf{1}} \right) \cr
& \int {\frac{x}{{{{\left( {4 + 5x} \right)}^2}}}} dx = \frac{1}{{{{\left( 5 \right)}^2}}}\left( {\frac{4}{{4 + \left( 5 \right)x}} + \ln \left| {4 + \left( 5 \right)x} \right|} \right) + C \cr
& {\text{Simplify}} \cr
& \int {\frac{x}{{{{\left( {4 + 5x} \right)}^2}}}} dx = \frac{1}{{25}}\left( {\frac{4}{{4 + 5x}} + \ln \left| {4 + 5x} \right|} \right) + C \cr
& = \frac{4}{{25\left( {4 + 5x} \right)}} + \frac{1}{{25}}\ln \left| {4 + 5x} \right| + C \cr} $$
