Answer
$$\frac{1}{5}\ln \left| {\frac{x}{{2x + 5}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{x\left( {2x + 5} \right)}}} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{Rewrite the integrand}} \cr
& = \int {\frac{1}{{x\left( {5 + 2x} \right)}}} dx \cr
& {\text{The integrand has a expression in the form }}a + bu{} \cr
& {\text{Use formula 65}} \cr
& \left( {65} \right):\,\,\,\,\,\int {\frac{{du}}{{u\left( {a + bu} \right)}}} = \frac{1}{a}\ln \left| {\frac{u}{{a + bu}}} \right| + C \cr
& {\text{let }}u = x,\,\,\,a = 5{\text{ and }}b = 2 \cr
& \int {\frac{1}{{x\left( {5 + 2x} \right)}}} dx = \frac{1}{5}\ln \left| {\frac{x}{{5 + 2x}}} \right| + C \cr
& or \cr
& = \frac{1}{5}\ln \left| {\frac{x}{{2x + 5}}} \right| + C \cr} $$
