Answer
$$\left( {4{x^2} + 10x} \right)\ln \left( {5x} \right) - 2{x^2} - 10x + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {8x + 10} \right)\ln \left( {5x} \right)} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = \ln \left( {5x} \right){\text{ then }}du = \frac{1}{x}dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \left( {8x + 10} \right)dx{\text{ then }}v = 4{x^2} + 10x \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\left( {8x + 10} \right)\ln \left( {5x} \right)} dx = \left( {4{x^2} + 10x} \right)\ln \left( {5x} \right) - \int {\left( {4{x^2} + 10x} \right)} \left( {\frac{1}{x}dx} \right) \cr
& \int {\left( {8x + 10} \right)\ln \left( {5x} \right)} dx = \left( {4{x^2} + 10x} \right)\ln \left( {5x} \right) - \int {\left( {4x + 10} \right)} dx \cr
& \int {\left( {8x + 10} \right)\ln \left( {5x} \right)} dx = \left( {4{x^2} + 10x} \right)\ln \left( {5x} \right) - \int {4x} dx - \int {10} dx \cr
& {\text{integrate}} \cr
& \int {\left( {8x + 10} \right)\ln \left( {5x} \right)} dx = \left( {4{x^2} + 10x} \right)\ln \left( {5x} \right) - 2{x^2} - 10x + C \cr} $$
