Answer
\[ = \ln \left|\frac{{{\,{{\left( {x - 1} \right)}^2}} }}{{{x^2} + 4x + 5}}\right| + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}}} dx \hfill \\
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Using\,\,partial\,fractions \hfill \\
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\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{A}{{x - 1}} + \frac{{Bx + C}}{{{x^2} + 4x + 5}} \hfill \\
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{\text{find}}\,\,{\text{the}}\,\,{\text{constants}} \hfill \\
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A = 2,{\text{ }}B = - 2,{\text{ C}} = - 10 \hfill \\
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\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{2}{{x - 1}} - \frac{{2x - 10}}{{{x^2} + 4x + 5}} \hfill \\
\hfill \\
{\text{Therefore}}{\text{,}} \hfill \\
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\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{2}{{x - 1}} - \frac{{2x + 4 - 14}}{{{x^2} + 4x + 4 + 1}} \hfill \\
\hfill \\
\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{2}{{x - 1}} - \frac{{2\,\left( {x + 2} \right) - 14}}{{\,{{\left( {x + 2} \right)}^2} + 1}} \hfill \\
\hfill \\
\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}} = \frac{2}{{x - 1}} - \frac{{2\,\left( {x + 2} \right)\,}}{{\,{{\left( {x + 2} \right)}^2} + 1}} - \frac{{14}}{{\,{{\left( {x + 2} \right)}^2} + 1}} \hfill \\
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substitute\,\,in\,\,\int_{}^{} {\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}}} dx \hfill \\
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\int_{}^{} {\frac{{20x}}{{\,\left( {x - 1} \right)\,\left( {{x^2} + 4x + 5} \right)}}dx} \,\, = \int_{}^{} {\,\left( {\frac{2}{{x - 1}} - \frac{{2\,\left( {x + 2} \right)\,}}{{\,{{\left( {x + 2} \right)}^2} + 1}} - \frac{{14}}{{\,{{\left( {x + 2} \right)}^2} + 1}}} \right)dx} \hfill \\
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let\,\,u = \,{\left( {x + 2} \right)^2} + 1 \hfill \\
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= 2\ln \left| {x - 1} \right| - \int_{}^{} {\frac{{du}}{u}} + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) \hfill \\
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{\text{integrate}}\,\,\,{\text{and}}\,\,{\text{simplify}} \hfill \\
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= \ln \left| {\,{{\left( {x - 1} \right)}^2}} \right| - \ln \left| u \right| + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) + C \hfill \\
\hfill \\
= \ln \left| {\,{{\left( {x - 1} \right)}^2}} \right| - \ln \left| {\,{{\left( {x + 2} \right)}^2} + 1} \right| + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) + C \hfill \\
\hfill \\
= \ln \left|\frac{{{\,{{\left( {x - 1} \right)}^2}} }}{{{x^2} + 4x + 5}}\right| + 14{\tan ^{ - 1}}\,\left( {x + 2} \right) + C \hfill \\
\hfill \\
\end{gathered} \]
