## Thomas' Calculus 13th Edition

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