Question Number 88422 by abdomathmax last updated on 10/Apr/20
$${calculate}\:\:\int_{\mathrm{1}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 12/Apr/20
$${first}\:\:{let}\:{find}\:{I}\:=\:\int\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${I}\:=\int\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left({x}−{i}\right)^{\mathrm{2}} \left({x}+{i}\right)^{\mathrm{2}} }\:=\int\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} \left(\frac{{x}−{i}}{{x}+{i}}\right)^{\mathrm{2}} \left({x}+{i}\right)^{\mathrm{4}} } \\ $$$${changement}\:\frac{{x}−{i}}{{x}+{i}}\:={t}\:{give}\:{x}−{i}\:={tx}+{it}\:\Rightarrow\left(\mathrm{1}−{t}\right){x}\:={i}\left(\mathrm{1}+{t}\right) \\ $$$${x}\:={i}\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\:\Rightarrow\frac{{dx}}{{dt}}\:={i}×\frac{\mathrm{1}−{t}+\left(\mathrm{1}+{t}\right)}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}{i}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${x}+\mathrm{1}\:=\frac{{i}+{it}}{\mathrm{1}−{t}}+\mathrm{1}\:=\frac{{i}+{it}+\mathrm{1}−{t}}{\mathrm{1}−{t}}\:=\frac{\left(−\mathrm{1}+{i}\right){t}+{i}+\mathrm{1}}{\mathrm{1}−{t}} \\ $$$${x}+{i}\:=\frac{{i}+{it}}{\mathrm{1}−{t}}+{i}\:=\frac{{i}+{it}+{i}−{it}}{\mathrm{1}−{t}}\:=\frac{\mathrm{2}{i}}{\mathrm{1}−{t}}\:\Rightarrow= \\ $$$${I}\:=\int\:\:\:\frac{\mathrm{2}{idt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left(\frac{\left({i}−\mathrm{1}\right){t}\:+{i}+\mathrm{1}}{\mathrm{1}−{t}}\right)^{\mathrm{3}} {t}^{\mathrm{2}} \left(\frac{\mathrm{2}{i}}{\mathrm{1}−{t}}\right)^{\mathrm{4}} } \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\int\:\:\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{7}} \:{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left\{\left({i}−\mathrm{1}\right){t}\:+{i}+\mathrm{1}\right\}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{i}}\:\int\:\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{5}} }{\left\{\left({i}−\mathrm{1}\right){t}\:+{i}+\mathrm{1}\right)^{\mathrm{3}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{8}{i}\left({i}−\mathrm{1}\right)^{\mathrm{3}} }\int\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{5}} }{\left({t}\:+\frac{{i}+\mathrm{1}}{{i}−\mathrm{1}}\right)^{\mathrm{3}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}{i}\left({i}−\mathrm{1}\right)^{\mathrm{3}} }\int\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{5}} }{\left({t}−{i}\right)^{\mathrm{3}} }{dt}\:\Rightarrow \\ $$$$\mathrm{8}{i}\left({i}−\mathrm{1}\right)^{\mathrm{3}} \:{I}\:=\int\:\:\frac{\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{C}_{\mathrm{5}} ^{{k}} \:{t}^{{k}} \left(−\mathrm{1}\right)^{\mathrm{5}−{k}} }{\left({t}−{i}\right)^{\mathrm{3}} }{dt} \\ $$$$=−\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{5}} ^{{k}} \:\:\int\:\frac{{t}^{{k}} }{\left({t}−{i}\right)^{\mathrm{3}} }{dt} \\ $$$$=−{C}_{\mathrm{5}} ^{\mathrm{0}} \:\int\:\frac{{dt}}{\left({t}−{i}\right)^{\mathrm{3}} }\:+{C}_{\mathrm{5}} ^{\mathrm{1}} \:\int\:\frac{{t}}{\left({t}−{i}\right)^{\mathrm{3}} }{dt}\:−{C}_{\mathrm{5}} ^{\mathrm{2}} \:\int\:\:\frac{{t}^{\mathrm{2}} }{\left({t}−{i}\right)^{\mathrm{3}} }{dt}+… \\ $$$$+{C}_{\mathrm{5}} ^{\mathrm{5}} \:\int\:\:\frac{{t}^{\mathrm{5}} }{\left({t}−{i}\right)^{\mathrm{3}} }{dt}\:…..{be}\:{continued}…. \\ $$$$ \\ $$