## Thomas' Calculus 13th Edition

$$3\ln \left| {\frac{{\root 3 \of x }}{{1 + \root 3 \of x }}} \right| + C$$
\eqalign{ & \int {\frac{{dx}}{{x\left( {1 + \root 3 \of x } \right)}}} \cr & {\text{Integrate by using the substitution method}} \cr & {\text{Let }}{u^3} = x,\,\,\,\,\,\,3{u^2}du = dx \cr & {\text{Write the integral in terms of }}u \cr & \int {\frac{{dx}}{{x\left( {1 + \root 3 \of x } \right)}}} = \int {\frac{{3{u^2}du}}{{{u^3}\left( {1 + \root 3 \of {{u^3}} } \right)}}} \cr & = \int {\frac{{3du}}{{u\left( {1 + u} \right)}}} \cr & = 3\int {\frac{{du}}{{u\left( {1 + u} \right)}}} \cr & \cr & {\text{Decompose the integrand into partial fractions}} \cr & \frac{1}{{u\left( {1 + u} \right)}} = \frac{A}{u} + \frac{B}{{1 + u}} \cr & {\text{multiply by }}u\left( {1 + u} \right) \cr & 1 = A\left( {1 + u} \right) + Bu \cr & \,\,\,{\text{if }}u = 0,\,\, then\,\,A = 1 \cr & \,\,\,{\text{if }}u = - 1,\,\, then\,\,B = - 1 \cr & \frac{1}{{u\left( {1 + u} \right)}} = \frac{A}{u} + \frac{B}{{1 + u}} = \frac{1}{u} - \frac{1}{{1 + u}} \cr & \cr & {\text{Then}} \cr & 3\int {\frac{{du}}{{u\left( {1 + u} \right)}}} = 3\int {\left( {\frac{1}{u} - \frac{1}{{1 + u}}} \right)} du \cr & {\text{integrating}} \cr & = 3\left( {\ln \left| u \right| - \ln \left| {1 + u} \right|} \right) + C \cr & = 3\ln \left| {\frac{u}{{1 + u}}} \right| + C \cr & \cr & {\text{Write the integral in terms of }}x,{\text{ replace }}\root 3 \of x {\text{ for }}u \cr & = 3\ln \left| {\frac{{\root 3 \of x }}{{1 + \root 3 \of x }}} \right| + C \cr}