Question Number 86013 by mathmax by abdo last updated on 26/Mar/20
$${calculate}\:\:\int_{\mathrm{1}+\sqrt{\mathrm{2}}} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{3}} } \\ $$
Commented by mathmax by abdo last updated on 28/Mar/20
$${I}\:=\int_{\mathrm{1}+\sqrt{\mathrm{2}}} ^{+\infty} \:\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{3}} }\:\Rightarrow{I}\:=\int_{\mathrm{1}+\sqrt{\mathrm{2}}} ^{+\infty} \:\frac{{dx}}{\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{6}} } \\ $$$${changement}\:\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}\:={t}\:\:{give}\:{x}−\mathrm{1}\:={tx}+\mathrm{2}{t}\:\Rightarrow\left(\mathrm{1}−{t}\right){x}=\mathrm{2}{t}+\mathrm{1}\:\Rightarrow \\ $$$${x}\:=\frac{\mathrm{2}{t}+\mathrm{1}}{\mathrm{1}−{t}}\:\Rightarrow\frac{{dx}}{{dt}}\:=\frac{\mathrm{2}\left(\mathrm{1}−{t}\right)−\left(\mathrm{2}{t}+\mathrm{1}\right)\left(−\mathrm{1}\right)}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}−\mathrm{2}{t}+\mathrm{2}{t}+\mathrm{1}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{3}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${x}+\mathrm{2}\:=\frac{\mathrm{2}{t}+\mathrm{1}}{\mathrm{1}−{t}}\:+\mathrm{2}\:=\frac{\mathrm{2}{t}+\mathrm{1}+\mathrm{2}−\mathrm{2}{t}}{\mathrm{1}−{t}}\:=\frac{\mathrm{3}}{\mathrm{1}−{t}}\:\Rightarrow \\ $$$${I}\:=\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{3}{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:{t}^{\mathrm{3}} \left(\frac{\mathrm{3}}{\mathrm{1}−{t}}\right)^{\mathrm{6}} }\:\:=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{6}} }{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:{t}^{\mathrm{3}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{4}} }{{t}^{\mathrm{3}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:\frac{\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:{C}_{\mathrm{4}} ^{{k}} \:{t}^{{k}} \left(−\mathrm{1}\right)^{\mathrm{4}−{k}} }{{t}^{\mathrm{3}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:\left(\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{4}} ^{{k}} \:{t}^{{k}−\mathrm{3}} \right)\:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }\sum_{{k}=\mathrm{0}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{4}} ^{{k}} \:\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:{t}^{{k}−\mathrm{3}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }\sum_{{k}=\mathrm{0}\:{and}\:{k}\neq\mathrm{2}} ^{{k}} \:\:\:\left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{4}} ^{{k}} \left[\frac{\mathrm{1}}{{k}−\mathrm{2}}{t}^{{k}−\mathrm{2}} \right]_{\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \:+{C}_{\mathrm{4}} ^{\mathrm{2}} \left[{ln}\mid{t}\mid\right]_{\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}} ^{\mathrm{1}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{5}} }\sum_{{k}=\mathrm{0}\:{and}\:{k}\neq\mathrm{2}} ^{{k}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{4}} ^{{k}} }{{k}−\mathrm{2}}\left\{\mathrm{1}−\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}\right)^{{k}−\mathrm{2}} \right\}−{C}_{\mathrm{4}} ^{\mathrm{2}} {ln}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}\right) \\ $$