find-x-2-x-2-1-dx-

Question Number 36424 by abdo.msup.com last updated on 01/Jun/18

$${find}\:\int\:\:{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$

Commented by prof Abdo imad last updated on 03/Jun/18

changement x = ch(t) give I = ∫ ch^2 t sh(t)sh(t)dt=∫ (ch(t)sh(t))^2 dt =(1/4) ∫ sh^2 (2t)dt =(1/8) ∫ (ch(4t)−1)dt = −(t/8) + (1/(32)) sh(4t) =−(1/8) argch(x) +(1/(32))sh{4argch(x)} =−(1/8)ln(x +(√(x^2 −1))) +(1/(32))sh{4ln(x +(√(x^2 −1)))} but sh{4ln(x+(√(x^2 −1)))=sh{ln(x+(√(x^2 −1)))^2 } = (((x+(√(x^2 −1)))^2 +(x+(√(x^2 −1)))^(−2) )/2) I = (1/(64)){ (x+(√(x^2 −1)))^2 +(x+(√(x^2 −1)))^(−2) } −(1/8)ln(x +(√(x^2 −1))) .

$${changement}\:{x}\:=\:{ch}\left({t}\right)\:{give}\: \\ $$$${I}\:=\:\int\:{ch}^{\mathrm{2}} {t}\:{sh}\left({t}\right){sh}\left({t}\right){dt}=\int\:\left({ch}\left({t}\right){sh}\left({t}\right)\right)^{\mathrm{2}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int\:\:{sh}^{\mathrm{2}} \left(\mathrm{2}{t}\right){dt}\:\:=\frac{\mathrm{1}}{\mathrm{8}}\:\int\:\:\left({ch}\left(\mathrm{4}{t}\right)−\mathrm{1}\right){dt} \\ $$$$=\:−\frac{{t}}{\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{32}}\:{sh}\left(\mathrm{4}{t}\right)\: \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}\:{argch}\left({x}\right)\:+\frac{\mathrm{1}}{\mathrm{32}}{sh}\left\{\mathrm{4}{argch}\left({x}\right)\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left({x}\:+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)\:+\frac{\mathrm{1}}{\mathrm{32}}{sh}\left\{\mathrm{4}{ln}\left({x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\right\} \\ $$$${but}\:{sh}\left\{\mathrm{4}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)={sh}\left\{{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \right\}\right. \\ $$$$=\:\frac{\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:+\left({x}+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)^{−\mathrm{2}} }{\mathrm{2}} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{64}}\left\{\:\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:+\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{−\mathrm{2}} \right\} \\ $$$$−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left({x}\:+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)\:. \\ $$

Commented by abdo mathsup last updated on 04/Jun/18

I = (1/(64)){ (x +(√(x^2 −1)))^2 −(x +(√(x^2 −1)))^(−2) } −(1/8)ln(x +(√(x^2 −1))) .

$${I}\:=\:\frac{\mathrm{1}}{\mathrm{64}}\left\{\:\left({x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:−\left({x}\:+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)^{−\mathrm{2}} \right\} \\ $$$$−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left({x}\:+\sqrt{{x}^{\mathrm{2}} \:−\mathrm{1}}\right)\:. \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply