## Calculus: Early Transcendentals (2nd Edition)

$= 1 - \ln 2$
$\begin{gathered} \int_1^\infty {\frac{{dx}}{{{x^2}\,\left( {x + 1} \right)}}} \hfill \\ \hfill \\ Integrating\,\,by\,\,partial\,\,fractions \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{{x^2}\,\left( {x + 1} \right)}}} \,\, = \,\,\int_{}^{} {\,\left( { - \frac{1}{x} + \frac{1}{{{x^2}}} + \frac{1}{{x - 1}}} \right)\,dx} \hfill \\ \hfill \\ = - \ln \left| x \right| - \frac{1}{x} + \ln \left| {x + 1} \right| + C \hfill \\ \hfill \\ = \ln \left| {\frac{{x + 1}}{x}} \right| - \frac{1}{x} + C \hfill \\ \hfill \\ use\,\,the\,\,Definition\,\,of\,\,improper\,\,integral\, \hfill \\ \hfill \\ \int_a^\infty {f\,\left( x \right)} \,dx = \,\,\,\,\,\mathop {\,\lim }\limits_{b \to \infty } \int_a^b {f\,\left( x \right)dx} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_1^\infty {\frac{{dx}}{{{x^2}\,\left( {x + 1} \right)}}} \,\, = \,\,\mathop {\,\lim }\limits_{b \to \infty } \,\,\int_1^b {\frac{{dx}}{{{x^2}\,\left( {x + 1} \right)}}} \hfill \\ \hfill \\ = \mathop {\,\lim }\limits_{b \to \infty } \,\left( {\,\left( {\ln \left| {1 + \frac{1}{b}} \right| - \frac{1}{b}} \right) - \,\left( {\ln \left| {1 + \frac{1}{1}} \right| - \frac{1}{1}} \right)} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \,\left( {\,\left( {\ln \left| {1 + 0} \right| - 0} \right) - \,\left( {\ln \left| 2 \right| - 1} \right)} \right) \hfill \\ \hfill \\ = 1 - \ln 2 \hfill \\ \hfill \\ \end{gathered}$