Answer
$$x\ln \left| {4x + 5} \right| - x + \frac{5}{4}\ln \left| {4x + 5} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\ln \left| {4x + 5} \right|} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = \ln \left| {4x + 5} \right|{\text{ then }}du = \frac{4}{{4x + 5}}dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = dx{\text{ then }}v = x \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\ln \left| {4x + 5} \right|} dx = \ln \left| {4x + 5} \right|\left( x \right) - \int {\left( x \right)\left( {\frac{4}{{4x + 5}}dx} \right)} \cr
& {\text{simplifying}} \cr
& \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - \int {\frac{{4x}}{{4x + 5}}} dx \cr
& {\text{use the long division for }}\frac{{4x}}{{4x + 5}}{\text{ to obtain }}1 - \frac{5}{{4x + 5}}.{\text{ then}} \cr
& \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - \int {\left( {1 - \frac{5}{{4x + 5}}} \right)} dx \cr
& \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - \int {dx} + \int {\frac{5}{{4x + 5}}} dx \cr
& {\text{integrating}} \cr
& \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - \int {dx} + \frac{5}{4}\int {\frac{4}{{4x + 5}}} dx3 \cr
& \int {\ln \left| {4x + 5} \right|} dx = x\ln \left| {4x + 5} \right| - x + \frac{5}{4}\ln \left| {4x + 5} \right| + C \cr} $$
