Chapter 8: Techniques of Integration - Practice Exercises - Page 519: 57

$$\ln 3$$

$$\eqalign{ & \int_3^\infty {\frac{{2du}}{{{u^2} - 2u}}} \cr & {\text{By the definition of the improper integrals, we have }} \cr & \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx \cr & {\text{then}}{\text{,}} \cr & \int_3^\infty {\frac{{2du}}{{{u^2} - 2u}}} = \mathop {\lim }\limits_{b \to \infty } \int_3^b {\frac{{2du}}{{{u^2} - 2u}}} \cr & \cr & {\text{Integrate }}\int {\frac{{2du}}{{{u^2} - 2u}}{\text{ by partial fractions method}}} \cr & \,\,\,\,\,\frac{2}{{{u^2} - 2u}} = \frac{2}{{u\left( {u - 2} \right)}} = \frac{A}{u} + \frac{B}{{u - 2}} \cr & \,\,\,\,\,2 = A\left( {u - 2} \right) + Bu\,\,\,\left( {\bf{1}} \right) \cr & \,\,\,\,\,{\text{let}}\,\,\,u = 0{\text{ into the equation }}\left( {{\bf{21}}} \right) \cr & \,\,\,\,\,2 = A\left( {0 - 2} \right) + B\left( 0 \right) \cr & \,\,\,\,\,A = - 1 \cr & \,\,\,\,{\text{let}}\,\,\,u = 2{\text{ into the equation }}\left( {{\bf{21}}} \right) \cr & \,\,\,\,\,2 = A\left( 0 \right) + B\left( 2 \right) \cr & \,\,\,\,\,B = 1 \cr & \,\,\,\,{\text{Then}}{\text{,}}\,\,\,\frac{2}{{u\left( {u - 2} \right)}} = \frac{A}{u} + \frac{B}{{u - 2}} = - \frac{1}{u} + \frac{1}{{u - 2}} \cr & \,\,\,\,\int {\frac{{2du}}{{{u^2} - 2u}} = - \int {\frac{1}{u}} du + \int {\frac{1}{{u - 2}}} } \cr & \,\,\,\,\int {\frac{{2du}}{{{u^2} - 2u}} = } - \ln \left| u \right| + \ln \left| {u - 2} \right| + C \cr & \,\,\,\,\int {\frac{{2du}}{{{u^2} - 2u}} = } \ln \left| {\frac{{u - 2}}{u}} \right| + C \cr & \,\,\,\,\int {\frac{{2du}}{{{u^2} - 2u}} = } \ln \left| {1 - \frac{2}{u}} \right| + C \cr & \cr & \mathop {\lim }\limits_{b \to \infty } \int_3^b {\frac{{2du}}{{{u^2} - 2u}}} = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {1 - \frac{2}{u}} \right|} \right)_3^b \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {1 - \frac{2}{b}} \right| - \ln \left| {1 - \frac{2}{3}} \right|} \right) \cr & = \mathop {\lim }\limits_{b \to \infty } \left( {\ln \left| {1 - \frac{2}{b}} \right| - \ln \frac{1}{3}} \right) \cr & {\text{Evaluate the limit when }}b \to \infty \cr & = \ln \left| {1 - \frac{2}{\infty }} \right| - \ln \frac{1}{3} \cr & = \ln \left| 1 \right| - \ln \frac{1}{3} \cr & = - \ln \frac{1}{3} \cr & = \ln 3 \cr} $$