Question Number 89146 by M±th+et£s last updated on 15/Apr/20
$$\left.\mathrm{1}\right)\int{x}\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}\:{dx} \\ $$$$\left.\mathrm{2}\right)\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$$$\left.\mathrm{3}\right)\int\sqrt{−{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}}\:{dx} \\ $$
Commented by mathmax by abdo last updated on 16/Apr/20
$$\left.\mathrm{1}\right)\:{I}\:=\int{x}\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}{dx}\:{changement}\:\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}={t}\:{give}\:\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}−\mathrm{2}={t}^{\mathrm{2}} {x}\:+{t}^{\mathrm{2}} \:\Rightarrow\left(\mathrm{1}−{t}^{\mathrm{2}} \right){x}\:={t}^{\mathrm{2}} \:+\mathrm{2}\:\Rightarrow{x}=\frac{{t}^{\mathrm{2}} \:+\mathrm{2}}{\mathrm{1}−{t}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{{dx}}{{dt}}\:=\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\mathrm{2}{t}\left({t}^{\mathrm{2}} \:+\mathrm{2}\right)}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{4}{t}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{6}{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{I}\:=\int\frac{{t}^{\mathrm{2}} \:+\mathrm{2}}{\mathrm{1}−{t}^{\mathrm{2}} }×{t}×\frac{\mathrm{6}{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt}\:=−\mathrm{6}\:\int\:\:\frac{{t}^{\mathrm{3}} \:+\mathrm{2}{t}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }{dt}\:\Rightarrow \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}{I}\:\:=\int\:\:\frac{{t}^{\mathrm{3}} \:+\mathrm{2}{t}}{\left({t}−\mathrm{1}\right)^{\mathrm{3}} \left({t}+\mathrm{1}\right)^{\mathrm{3}} }{dt}\:=\int\:\:\frac{{t}^{\mathrm{3}} \:+\mathrm{2}{t}}{\left(\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\right)^{\mathrm{3}} \left({t}+\mathrm{1}\right)^{\mathrm{6}} }\:{we}\:{do}\:{the}\:{changement} \\ $$$$\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\:={u}\:\Rightarrow{t}−\mathrm{1}\:={ut}+{u}\:\Rightarrow\left(\mathrm{1}−{u}\right){t}\:={u}+\mathrm{1}\:\Rightarrow{t}\:=\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\:\Rightarrow \\ $$$$\frac{{dt}}{{du}}\:=\frac{\mathrm{1}−{u}+\mathrm{1}+{u}}{\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} }\:\:{also}\:{t}+\mathrm{1}\:=\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}\:+\mathrm{1}\:=\frac{\mathrm{1}+{u}+\mathrm{1}−{u}}{\mathrm{1}−{u}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}−{u}}\:\Rightarrow \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}{I}\:=\:\int\:\:\:\frac{\frac{\left(\mathrm{1}+{u}\right)^{\mathrm{3}} }{\left(\mathrm{1}−{u}\right)^{\mathrm{3}} }+\mathrm{2}\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}}{{u}^{\mathrm{3}} \left(\frac{\mathrm{2}}{\mathrm{1}−{u}}\right)^{\mathrm{6}} }×\frac{\mathrm{2}{du}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }\:\int\:\:\:\:\frac{\left(\mathrm{1}+{u}\right)^{\mathrm{3}} \:+\mathrm{2}\left(\mathrm{1}+{u}\right)\left(\mathrm{1}−{u}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{u}\right)^{\mathrm{3}} \:{u}^{\mathrm{3}} }×\left({u}−\mathrm{1}\right)^{\mathrm{4}} \:{du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{5}} }\int\:\:\:\frac{\left({u}−\mathrm{1}\right)\left\{\:{u}^{\mathrm{3}} \:+\mathrm{3}{u}^{\mathrm{2}} \:+\mathrm{3}{u}\:+\mathrm{1}\:+\mathrm{2}\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} −\mathrm{2}{u}\:+\mathrm{1}\right)\right\}}{{u}^{\mathrm{3}} }{du} \\ $$$$\Rightarrow{I}\:=\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{5}} }\:\int\:\frac{\left({u}−\mathrm{1}\right)\left\{\:{u}^{\mathrm{3}} \:+\mathrm{3}{u}^{\mathrm{2}} \:+\mathrm{3}{u}\:+\mathrm{1}\:+\mathrm{2}\left({u}^{\mathrm{3}} −\mathrm{2}{u}^{\mathrm{2}} +{u}+{u}^{\mathrm{2}} −\mathrm{2}{u}+\mathrm{1}\right)\right\}}{{u}^{\mathrm{3}} }{du} \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}}\int\:\frac{\left({u}−\mathrm{1}\right)\left\{{u}^{\mathrm{3}} \:+\mathrm{3}{u}^{\mathrm{2}} \:+\mathrm{3}{u}+\mathrm{1}\:+\left(\mathrm{2}{u}^{\mathrm{3}} −\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}{u}+\mathrm{2}\right)\right\}}{{u}^{\mathrm{3}} }{du} \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}}\int\:\:\frac{\left({u}−\mathrm{1}\right)\left\{\mathrm{5}{u}^{\mathrm{3}} +{u}^{\mathrm{2}} +{u}\:+\mathrm{3}\right\}}{{u}^{\mathrm{3}} }{du} \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}}\int\:\frac{\mathrm{5}{u}^{\mathrm{4}} \:+{u}^{\mathrm{3}} +{u}^{\mathrm{2}} +\mathrm{3}{u}\:−\mathrm{5}{u}^{\mathrm{3}} −{u}^{\mathrm{2}} −{u}−\mathrm{3}}{{u}^{\mathrm{3}} }\:{du} \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}}\int\:\frac{\mathrm{5}{u}^{\mathrm{4}} −\mathrm{4}{u}^{\mathrm{3}} +\mathrm{2}{u}−\mathrm{3}}{{u}^{\mathrm{3}} }{du} \\ $$$$=\frac{\mathrm{15}}{\mathrm{16}}\:\int{u}\:{du}\:−\frac{\mathrm{3}}{\mathrm{4}}\:\int\:{du}\:\:+\frac{\mathrm{3}}{\mathrm{8}}\:\int\:\frac{{du}}{{u}^{\mathrm{2}} }−\frac{\mathrm{9}}{\mathrm{16}}\int\:\frac{{du}}{{u}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{15}}{\mathrm{32}}{u}^{\mathrm{2}} \:−\frac{\mathrm{3}}{\mathrm{4}}{u}\:\:−\frac{\mathrm{3}}{\mathrm{8}{u}}\:+\frac{\mathrm{9}}{\mathrm{32}{u}^{\mathrm{2}} }\:+{C} \\ $$$$=\frac{\mathrm{15}}{\mathrm{32}}\left(\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\right)^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\right)\:−\frac{\mathrm{3}}{\mathrm{8}}\left(\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\right)\:+\frac{\mathrm{9}}{\mathrm{32}}\left(\frac{{t}+\mathrm{1}}{{t}−\mathrm{1}}\right)^{\mathrm{2}} \:+{C} \\ $$$$=\frac{\mathrm{15}}{\mathrm{32}}\left(\frac{\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}−\mathrm{1}}{\:\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}+\mathrm{1}}}\right)^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}−\mathrm{1}}{\:\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}+\mathrm{1}}\right)−\frac{\mathrm{3}}{\mathrm{8}}\left(\frac{\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}+\mathrm{1}}{\:\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}−\mathrm{1}}\right) \\ $$$$+\frac{\mathrm{9}}{\mathrm{32}}\left(\frac{\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}+\mathrm{1}}{\:\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}−\mathrm{1}}}\right)^{\mathrm{2}} \:+{C} \\ $$
Commented by mathmax by abdo last updated on 16/Apr/20
$${error}\:{at}\:{line}\:\mathrm{4}\:\:\:{I}\:=−\mathrm{6}\:\int\:\:\frac{{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }{dt}\:\:{but}\:{give}\:{opprtonity}\:{to} \\ $$$${the}\:{method}\:…. \\ $$
Commented by M±th+et£s last updated on 16/Apr/20
$${thank}\:{you}\:{for}\:{the}\:{try}\:{sir} \\ $$
Commented by mathmax by abdo last updated on 16/Apr/20
$$\left.\mathrm{2}\right)\:{I}\:=\int\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}\:\Rightarrow\:{I}=\int\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$${we}\:{do}\:{the}\:{changement}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{sh}\left({t}\right)\:\Rightarrow \\ $$$${I}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\int\frac{{ch}\left({t}\right){dt}}{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sht}+\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({t}\right)}\:=\int\:\frac{\mathrm{2}{dt}}{\mathrm{1}+\sqrt{\mathrm{3}}{sh}\left({t}\right)}\:=\int\frac{\mathrm{2}{dt}}{\mathrm{1}+\sqrt{\mathrm{3}}\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}} \\ $$$$=\mathrm{4}\:\int\:\:\frac{{dt}}{\mathrm{2}\:+\sqrt{\mathrm{3}}{e}^{{t}} −\sqrt{\mathrm{3}}{e}^{−{t}} }\:=_{{e}^{{t}} ={z}} \:\mathrm{4}\:\int\:\:\frac{{dz}}{{z}\left(\mathrm{2}+\sqrt{\mathrm{3}}{z}\:−\sqrt{\mathrm{3}}{z}^{−\mathrm{1}} \right)} \\ $$$$=\mathrm{4}\:\int\:\:\frac{{dz}}{\mathrm{2}{z}+\sqrt{\mathrm{3}}{z}^{\mathrm{2}} −\sqrt{\mathrm{3}}}\:=\mathrm{4}\:\int\:\frac{{dz}}{\:\sqrt{\mathrm{3}}{z}^{\mathrm{2}} \:+\mathrm{2}{z}−\sqrt{\mathrm{3}}} \\ $$$$\Delta^{'} =\mathrm{1}+\mathrm{3}\:=\mathrm{4}\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\:{and}\:{z}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:=−\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$${I}\:=\mathrm{4}\:\int\:\frac{{dz}}{\:\sqrt{\mathrm{3}}\left({z}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\left({z}+\sqrt{\mathrm{3}}\right)}\:=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)}\int\:\left(\frac{\mathrm{1}}{{z}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}−\frac{\mathrm{1}}{{z}+\sqrt{\mathrm{3}}}\right){dz} \\ $$$$={ln}\mid\frac{{z}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{{z}+\sqrt{\mathrm{3}}}\mid\:+{C} \\ $$$$={ln}\mid\frac{\sqrt{\mathrm{3}}{z}−\mathrm{1}}{\:\sqrt{\mathrm{3}}{z}\:+\mathrm{3}}\mid\:+{C}\:={ln}\mid\frac{\sqrt{\mathrm{3}}{e}^{{t}} −\mathrm{1}}{\:\sqrt{\mathrm{3}}{e}^{{t}} \:+\mathrm{3}}\mid\:+{C}\:\:{we}\:{have}\:{t}={argsh}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$={ln}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${I}\:={ln}\mid\frac{\sqrt{\mathrm{3}}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}\:}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }−\mathrm{1}\right.}{\:\sqrt{\mathrm{3}}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }+\mathrm{3}\right.}\mid\:+{C} \\ $$
Answered by TANMAY PANACEA. last updated on 15/Apr/20
$$\left.\mathrm{3}\right)−{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10} \\ $$$$=−\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}−\mathrm{14}\right) \\ $$$$=\mathrm{14}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$${so}\:\int\sqrt{\mathrm{14}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:\:{dx} \\ $$$$\frac{\left({x}−\mathrm{2}\right)\sqrt{\mathrm{14}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} }}{\mathrm{2}}+\frac{\mathrm{14}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}−\mathrm{2}}{\:\sqrt{\mathrm{14}}}\right)+{c} \\ $$
Commented by peter frank last updated on 15/Apr/20
$${good} \\ $$
Commented by peter frank last updated on 15/Apr/20
$${good} \\ $$
Commented by TANMAY PANACEA. last updated on 15/Apr/20
$${most}\:{welcome}\:{sir} \\ $$
Commented by M±th+et£s last updated on 15/Apr/20
$${nice}\:{work}\:{thanx}\:{sir} \\ $$
Answered by TANMAY PANACEA. last updated on 15/Apr/20
$$\left.\mathrm{2}\right)\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}} \\ $$$${x}+\mathrm{1}=\frac{\mathrm{1}}{{t}} \\ $$$$\int\frac{−{dt}}{{t}^{\mathrm{2}} ×\frac{\mathrm{1}}{{t}}\sqrt{\left(\frac{\mathrm{1}}{{t}}−\mathrm{1}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{t}}−\mathrm{1}\right)+\mathrm{1}}} \\ $$$$\int\frac{−{dt}}{{t}\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\frac{\mathrm{2}}{{t}}+\mathrm{1}+\frac{\mathrm{1}}{{t}}}} \\ $$$$\int\frac{−{tdt}}{{t}\sqrt{\mathrm{1}−\mathrm{2}{t}+{t}^{\mathrm{2}} +{t}}} \\ $$$$\int\frac{−{dt}}{\:\sqrt{{t}^{\mathrm{2}} −{t}+\mathrm{1}}} \\ $$$$\int\frac{−{dt}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{2}.{t}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$\int\frac{−{dt}}{\:\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$=\left(−\mathrm{1}\right)\left[\frac{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+\frac{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}{ln}\left\{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\right]\right. \\ $$$${pls}\:{check}\:{and}\:{put}\:{t}=\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$
Commented by TANMAY PANACEA. last updated on 15/Apr/20
$${most}\:{welcome} \\ $$
Commented by M±th+et£s last updated on 15/Apr/20
$${its}\:{correct}\:{thank}\:{you}\:{sir} \\ $$
Commented by peter frank last updated on 20/Apr/20
$${help}\:{Qn}\:\mathrm{89833}\left({physics}\right) \\ $$$$ \\ $$
Answered by MJS last updated on 15/Apr/20
$$\int{x}\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}}\:\rightarrow\:{dx}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}\left({x}+\mathrm{1}\right){dt}\right] \\ $$$$=−\mathrm{6}\int\frac{{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=\frac{\mathrm{3}{t}\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{4}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{4}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{1}}= \\ $$$$=\frac{\mathrm{3}{t}\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{4}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{8}}\mathrm{ln}\:\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}= \\ $$$$=\frac{\left(\mathrm{2}{x}−\mathrm{5}\right)\sqrt{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{8}}\mathrm{ln}\:\mid\mathrm{2}{x}−\mathrm{1}−\mathrm{2}\sqrt{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}\mid\:+{C} \\ $$
Commented by M±th+et£s last updated on 15/Apr/20
$${thank}\:{you}\:{sir} \\ $$
Answered by TANMAY PANACEA. last updated on 15/Apr/20
$$\left.\mathrm{1}\right)\int\frac{{x}\left({x}−\mathrm{2}\right)}{\:\sqrt{\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)}}{dx} \\ $$$$\int\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+{x}−\mathrm{2}}}{dx} \\ $$$$\int\frac{{x}^{\mathrm{2}} −{x}−\mathrm{2}+\mathrm{2}−{x}}{\:\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{2}}}{dx} \\ $$$$\int\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{2}}\:{dx}+\int\frac{\mathrm{2}−{x}}{\:\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{2}}}{dx} \\ $$$$\int\sqrt{{x}^{\mathrm{2}} −\mathrm{2}.{x}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{9}}{\mathrm{4}}}\:{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}−\mathrm{3}}{\:\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{2}}}{dx} \\ $$$$\int\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }\:{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({x}^{\mathrm{2}} −{x}−\mathrm{2}\right)}{\:\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{2}}}{dx}+\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{dx}}{\:\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }\:} \\ $$$${I}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{I}_{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{I}_{\mathrm{3}} \\ $$$${I}_{\mathrm{1}} =\frac{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}}\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }\:−\frac{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}}{ln}\left\{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\left[\right.\right. \\ $$$$ \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{2}}\: \\ $$$${I}_{\mathrm{3}} ={ln}\left\{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}}\:\right\} \\ $$$${pls}\:{check} \\ $$
Commented by M±th+et£s last updated on 15/Apr/20
$${thank}\:{you}\:{verry}\:{much} \\ $$
Commented by TANMAY PANACEA. last updated on 15/Apr/20
$${most}\:{welcome}\:{sir} \\ $$
Answered by MJS last updated on 15/Apr/20
$$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2}{x}+\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} +\mathrm{3}}}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{3}}\right)\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{3}\left({t}^{\mathrm{2}} +\mathrm{3}\right)}}{{t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{3}}}{du}\right] \\ $$$$=\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\int\frac{{du}}{{u}^{\mathrm{2}} +\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}{u}−\mathrm{1}}= \\ $$$$=\int\frac{{du}}{{u}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}−\int\frac{{du}}{{u}+\sqrt{\mathrm{3}}}=\mathrm{ln}\:\left({u}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)\:−\mathrm{ln}\:\left({u}+\sqrt{\mathrm{3}}\right)\:= \\ $$$$=\mathrm{ln}\:\frac{{u}−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}}{{u}+\sqrt{\mathrm{3}}}\:=…=\mathrm{ln}\:\frac{{t}−\mathrm{3}+\mathrm{2}\sqrt{{t}^{\mathrm{2}} +\mathrm{3}}}{{t}+\mathrm{1}}\:= \\ $$$$=\mathrm{ln}\:\frac{{x}−\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{{x}+\mathrm{1}}\:+{C} \\ $$
Answered by 242242864 last updated on 16/Apr/20
$$ \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\:\:\int\sqrt{−{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10}\:}\:{dx} \\ $$$${consider}\:−{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{10} \\ $$$$=−\left[{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{10}\right] \\ $$$$−\left[\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{10}−\mathrm{4}\right] \\ $$$$−\left[\left({x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{14}\right] \\ $$$$\mathrm{14}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\int\sqrt{−{x}+\mathrm{4}{x}+\mathrm{10}}\:=\:\int\sqrt{\mathrm{14}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$\boldsymbol{{let}}\:\sqrt{\mathrm{14}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:=\sqrt{\boldsymbol{{a}}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta}\right)} \\ $$$$\mathrm{14}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\:\boldsymbol{{a}}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta}\right) \\ $$$$ \\ $$$$\mathrm{14}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:=\:\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \boldsymbol{\theta} \\ $$$$\Rightarrow\boldsymbol{{a}}^{\mathrm{2}} \:=\:\mathrm{14} \\ $$$$\left({x}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{14sin}\:^{\mathrm{2}} \theta \\ $$$$\left({x}−\mathrm{2}\right)=\sqrt{\mathrm{14}}\mathrm{sin}\:\theta \\ $$$$\mathrm{sin}\:\theta=\:\frac{\left({x}−\mathrm{2}\right)}{\:\sqrt{\mathrm{14}}} \\ $$$$\theta=\:\mathrm{sin}\:^{−\mathrm{1}\:} \frac{\left({x}−\mathrm{2}\right)}{\:\sqrt{\mathrm{14}}} \\ $$$${from}\:\left({x}−\mathrm{2}\right)=\:\sqrt{\mathrm{14}}\mathrm{sin}\:\theta \\ $$$${dx}=\sqrt{\mathrm{14}}\mathrm{cos}\:\theta{d}\theta \\ $$$$\int\sqrt{\mathrm{14}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \theta\right)}\:.\frac{\sqrt{\mathrm{14}}\mathrm{cos}\:\theta{d}\theta}{\mathrm{1}} \\ $$$$\int{d}\theta\:=\:\theta+{C} \\ $$$$=\mathrm{sin}\:^{−\mathrm{1}} \:\frac{\left({x}−\mathrm{2}\right)}{\:\sqrt{\mathrm{14}}}+{C} \\ $$$$ \\ $$$${something}\:{missing}? \\ $$$$ \\ $$