Question Number 40144 by maxmathsup by imad last updated on 16/Jul/18
$${find}\:\:\int_{\mathrm{1}} ^{\mathrm{2}} {x}\sqrt{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:+\mathrm{5}}\:{dx} \\ $$
Answered by maxmathsup by imad last updated on 19/Jul/18
![let I = ∫_1 ^2 x(√(x^2 −2x+5))dx I = ∫_1 ^2 x(√((x−1)^2 +4))dx changement x−1=2sh(t) give I = ∫_0 ^(argsh((1/2))) (1+2sh(t))2 ch(t) 2ch(t)dt =4 ∫_0 ^(argsh((1/2))) (1+2sh(t))ch^2 t dt =4 ∫_0 ^(argsh((1/2))) ch^2 t dt +8 ∫_0 ^(argsh((1/2))) sht ch^2 t dt =2 ∫_0 ^(ln((1/2)+(√(5/4)))) {ch(2t)−1}dt +(8/3) [ch^3 t]_0 ^(ln((1/2)+((√5)/2))) =−2ln((1/2) +((√5)/2)) +[sh(2t)]_0 ^(ln((1/2) +((√5)/2))) +(8/3){ (((((1/2)+((√5)/2))^3 −((1/2)+((√5)/2))^(−1) )/2))^3 −1} I=−2ln(((1+(√5))/2)) +{(((((1+(√5))/2))^2 −(((1+(√5))/2))^(−2) )/2)}+(1/3){(((1+(√5))/2))^3 −(((1+(√5))/2))^(−3) }](https://www.tinkutara.com/question/Q40339.png)
$${let}\:{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} {x}\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}}{dx} \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} {x}\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}}{dx}\:\:{changement}\:{x}−\mathrm{1}=\mathrm{2}{sh}\left({t}\right)\:{give} \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{{argsh}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \left(\mathrm{1}+\mathrm{2}{sh}\left({t}\right)\right)\mathrm{2}\:{ch}\left({t}\right)\:\mathrm{2}{ch}\left({t}\right){dt} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{{argsh}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\left(\mathrm{1}+\mathrm{2}{sh}\left({t}\right)\right){ch}^{\mathrm{2}} {t}\:{dt} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{{argsh}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:{ch}^{\mathrm{2}} {t}\:{dt}\:+\mathrm{8}\:\int_{\mathrm{0}} ^{{argsh}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:{sht}\:{ch}^{\mathrm{2}} {t}\:{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}\right)} \left\{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}\right\}{dt}\:\:+\frac{\mathrm{8}}{\mathrm{3}}\:\left[{ch}^{\mathrm{3}} {t}\right]_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)} \\ $$$$=−\mathrm{2}{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:+\left[{sh}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)} \:\:\:+\frac{\mathrm{8}}{\mathrm{3}}\left\{\:\left(\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{3}} −\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{−\mathrm{1}} }{\mathrm{2}}\right)^{\mathrm{3}} −\mathrm{1}\right\} \\ $$$${I}=−\mathrm{2}{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)\:+\left\{\frac{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{−\mathrm{2}} }{\mathrm{2}}\right\}+\frac{\mathrm{1}}{\mathrm{3}}\left\{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{3}} \:−\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{−\mathrm{3}} \right\} \\ $$
Commented by maxmathsup by imad last updated on 19/Jul/18
$${I}\:=…….+\frac{\mathrm{1}}{\mathrm{3}}\left\{\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{3}} \:+\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{−\mathrm{3}} \:−\mathrm{8}\right\} \\ $$