## Calculus: Early Transcendentals (2nd Edition)

$= \ln x - \frac{1}{{10}}\,\left( {1 + {x^{10}}} \right) + C$
$\begin{gathered} \int_{}^{} {\frac{{dx}}{{x\,\left( {{x^{10}} + 1} \right)}}} \hfill \\ \hfill \\ \,change\,\,the\,variables \hfill \\ \hfill \\ u = {x^5}\,\,\,\, \to \,\,\,\,\,du = 5{x^4}dx\,\,\, \to \,\,\,\,\frac{{du}}{{5u}} = \frac{{dx}}{x} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{x\,\left( {{x^{10}} + 1} \right)}}} = \int_{}^{} {\frac{1}{{\,{{\left( {{5^5}} \right)}^2} + 1}}} \,\,\,\left( {\frac{{dx}}{x}} \right) \hfill \\ \hfill \\ = \int_{}^{} {\frac{1}{{{u^2} + 1}}} \,\left( {\frac{{du}}{{5u}}} \right) = \frac{1}{5}\int_{}^{} {\frac{1}{{u\,\left( {{u^2} + 1} \right)}}\,du} \hfill \\ \hfill \\ Using\,\,integration\,\,formula \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{x\,\left( {{a^2} + {x^2}} \right)}} = \frac{1}{{2{a^2}}}\ln \,\left( {\frac{{{x^2}}}{{{a^2} + {x^2}}}} \right) + C} \hfill \\ \hfill \\ substituting\,\,\,1\,\,for\,\,{a^2} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{x\,\left( {{x^{10}} + 1} \right)}}} = \frac{1}{5}\int_{}^{} {\frac{1}{{u\,\left( {{u^2} + 1} \right)}}du} \hfill \\ \hfill \\ = \frac{1}{5}\,\left( {\frac{1}{{2\,\left( 1 \right)}}\ln \frac{{{u^2}}}{{1 + {u^2}}}} \right) + C \hfill \\ \hfill \\ = \frac{1}{{10}}\ln \frac{{{u^2}}}{{1 + {u^2}}} + C = \frac{1}{{10}}\,\ln \frac{{{x^{10}}}}{{1 + {x^{10}}}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{x\,\left( {{x^{10}} + 1} \right)}}} = \frac{1}{{10}}\,\left( {10\ln x - \ln \,\left( {1 + {x^{10}}} \right)} \right) + C \hfill \\ \hfill \\ Simplify \hfill \\ \hfill \\ = \ln x - \frac{1}{{10}}\,\left( {1 + {x^{10}}} \right) + C \hfill \\ \end{gathered}$