Question Number 67907 by A8;15: last updated on 02/Sep/19
Commented by mathmax by abdo last updated on 02/Sep/19
$${let}\:{I}\:=\int\:\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{6}}} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{6}=\mathrm{0}\rightarrow\Delta=\mathrm{1}−\mathrm{4}\left(−\mathrm{6}\right)\:=\mathrm{25}\:\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{5}}{\mathrm{2}}=\mathrm{2}\:\:{and}\: \\ $$$${x}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{5}}{\mathrm{2}}\:=−\mathrm{3}\:\Rightarrow{I}\:=\int\:\:\:\frac{{dx}}{{x}\sqrt{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)}}\:\:{we}\:{do}\:{the}\:{changement} \\ $$$$\sqrt{{x}−\mathrm{2}}={t}\:\Rightarrow{x}−\mathrm{2}={t}^{\mathrm{2}} \:\Rightarrow{x}={t}^{\mathrm{2}} \:+\mathrm{2}\:\Rightarrow{I}\:=\int\:\:\:\frac{\mathrm{2}{tdt}}{\left({t}^{\mathrm{2}} \:+\mathrm{2}\right){t}\sqrt{{t}^{\mathrm{2}} +\mathrm{2}+\mathrm{3}}} \\ $$$$=\int\:\:\:\frac{\mathrm{2}{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{2}\right)\sqrt{{t}^{\mathrm{2}} \:+\mathrm{5}}}\:\:{af}\:{ter}\:{we}\:{do}\:{the}\:{changement}\:{t}=\sqrt{\mathrm{5}}{sh}\left({u}\right)\:\Rightarrow \\ $$$${I}\:=\int\:\:\:\:\frac{\mathrm{2}\sqrt{\mathrm{5}}{ch}\left({u}\right)}{\left(\mathrm{25}{sh}^{\mathrm{2}} {u}\:+\mathrm{2}\right)\sqrt{\mathrm{5}}{ch}\left({u}\right)}{du}\:=\int\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{25}\frac{{ch}\left(\mathrm{2}{u}\right)−\mathrm{1}}{\mathrm{2}}+\mathrm{2}} \\ $$$$=\int\:\:\:\:\:\frac{\mathrm{4}{du}}{\mathrm{25}{ch}\left(\mathrm{2}{u}\right)−\mathrm{25}+\mathrm{4}}\:=\int\:\:\:\frac{\mathrm{4}{du}}{\mathrm{25}{ch}\left(\mathrm{2}{u}\right)−\mathrm{21}} \\ $$$$=\int\:\:\:\frac{\mathrm{4}{du}}{\mathrm{25}\frac{{e}^{\mathrm{2}{u}} +{e}^{−\mathrm{2}{u}} }{\mathrm{2}}−\mathrm{21}}\:=\int\:\:\frac{\mathrm{4}{du}}{\mathrm{25}{e}^{\mathrm{2}{u}} +\mathrm{25}{e}^{−\mathrm{2}{u}} −\mathrm{42}} \\ $$$$=_{{e}^{\mathrm{2}{u}} ={t}} \:\:\:\:\int\:\:\:\frac{\mathrm{4}}{\mathrm{25}{t}+\mathrm{25}{t}^{−\mathrm{1}} −\mathrm{42}}×\frac{{dt}}{\mathrm{2}{t}}\:=\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{25}{t}^{\mathrm{2}} +\mathrm{25}}\:=\frac{\mathrm{2}}{\mathrm{25}}\:\int\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{25}}\:{arctan}\left({t}\right)+{c}\:=\frac{\mathrm{2}}{\mathrm{25}}\:{arctan}\left({e}^{\mathrm{2}{u}} \right)\:+{c} \\ $$$${but}\:{u}={argsh}\left(\frac{{t}}{\:\sqrt{\mathrm{5}}}\right)\:={ln}\left(\frac{{t}}{\:\sqrt{\mathrm{5}}}+\sqrt{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\mathrm{5}}}\right)\:\Rightarrow\: \\ $$$${e}^{\mathrm{2}{u}} =\left(\frac{{t}}{\:\sqrt{\mathrm{5}}}+\sqrt{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\mathrm{5}}}\right)^{\mathrm{2}} \:\Rightarrow\:{I}\:=\frac{\mathrm{2}}{\mathrm{25}}\:{arctan}\left\{\left(\frac{{t}+\sqrt{\mathrm{5}+{t}^{\mathrm{2}} }}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} \right\}\:+{C} \\ $$
Answered by MJS last updated on 02/Sep/19
$$\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{6}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{12}−{x}}{\mathrm{5}{x}}\:\rightarrow\:{dx}=−\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{12}}{dt}\right] \\ $$$$=−\frac{\sqrt{\mathrm{6}}}{\mathrm{6}}\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=−\frac{\sqrt{\mathrm{6}}}{\mathrm{6}}\mathrm{arcsin}\:{t}\:= \\ $$$$=\frac{\sqrt{\mathrm{6}}}{\mathrm{6}}\mathrm{arcsin}\:\frac{{x}−\mathrm{12}}{\mathrm{5}{x}} \\ $$
Commented by MJS last updated on 02/Sep/19
$$\int\frac{{dx}}{{x}\sqrt{\left({x}+{a}^{\mathrm{2}} \right)\left({x}−{b}^{\mathrm{2}} \right)}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right){x}−\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){x}}\:\rightarrow\:{dx}=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){x}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{dt}\right] \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }}\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$