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Question Number 36417 by abdo.msup.com last updated on 01/Jun/18

calculate ∫_2 ^6 (dx/((√(x+1)) +(√(x−1))))

$${calculate}\:\int_{\mathrm{2}} ^{\mathrm{6}} \:\:\:\frac{{dx}}{\sqrt{{x}+\mathrm{1}}\:+\sqrt{{x}−\mathrm{1}}} \\ $$

Commented by abdo mathsup 649 cc last updated on 03/Jun/18

I = ∫_2 ^6 (((√(x+1)) −(√(x−1)))/2)dx ⇒ 2I = ∫_2 ^6 (√(x+1))dx −∫_2 ^6 (√(x−1))dx but changement (√(x+1)) =t give ∫_2 ^6 (√(x+1)) dx = ∫_(√3) ^(√7) t 2t dt = 2 ∫_(√3) ^(√7) t^2 dt = (2/3) { ((√7))^3 −((√3))^3 } and the changement (√(x−1))=t give ∫_2 ^6 (√(x−1))dx = ∫_1 ^(√5) t (2t)dt = (2/3){ ((√5))^3 −1} ⇒ I = (1/3){ 7(√7) −3(√3) −5(√5) +1}

$${I}\:=\:\int_{\mathrm{2}} ^{\mathrm{6}} \:\:\:\frac{\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}}{\mathrm{2}}{dx}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\:\int_{\mathrm{2}} ^{\mathrm{6}} \:\sqrt{{x}+\mathrm{1}}{dx}\:−\int_{\mathrm{2}} ^{\mathrm{6}} \:\sqrt{{x}−\mathrm{1}}{dx}\:\:{but}\:{changement} \\ $$$$\sqrt{{x}+\mathrm{1}}\:={t}\:{give}\:\int_{\mathrm{2}} ^{\mathrm{6}} \:\sqrt{{x}+\mathrm{1}}\:{dx}\:=\:\int_{\sqrt{\mathrm{3}}} ^{\sqrt{\mathrm{7}}} {t}\:\mathrm{2}{t}\:{dt} \\ $$$$=\:\mathrm{2}\:\int_{\sqrt{\mathrm{3}}} ^{\sqrt{\mathrm{7}}} \:{t}^{\mathrm{2}} \:{dt}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\left\{\:\left(\sqrt{\mathrm{7}}\right)^{\mathrm{3}} \:−\left(\sqrt{\mathrm{3}}\right)^{\mathrm{3}} \right\}\:{and}\:{the} \\ $$$${changement}\:\sqrt{{x}−\mathrm{1}}={t}\:{give} \\ $$$$\int_{\mathrm{2}} ^{\mathrm{6}} \:\sqrt{{x}−\mathrm{1}}{dx}\:=\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} {t}\:\left(\mathrm{2}{t}\right){dt}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\left\{\:\:\left(\sqrt{\mathrm{5}}\right)^{\mathrm{3}} \:−\mathrm{1}\right\}\:\Rightarrow \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left\{\:\mathrm{7}\sqrt{\mathrm{7}}\:−\mathrm{3}\sqrt{\mathrm{3}}\:\:−\mathrm{5}\sqrt{\mathrm{5}}\:\:+\mathrm{1}\right\} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

∫_2 ^6 (((√(x+1)) −(√(x−1)) )/2) =(1/2)∫_2 ^6 (√(x+1)) dx−(1/2)∫_2 ^6 (√(x−1)) dx x+1=t_1 ^2 x−1=t_2 ^2 so dx=2t_2 dt_2 dx=2t_1 dt_1 =(1/2)∫_((√3) ) ^((√7) ) t_1 ×2t_1 dt_1 −(1/2)∫_1 ^(√5) t_2 ×2t_2 dt_2 nw simple to get result...

$$\int_{\mathrm{2}} ^{\mathrm{6}} \frac{\sqrt{{x}+\mathrm{1}}\:−\sqrt{{x}−\mathrm{1}}\:}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{2}} ^{\mathrm{6}} \sqrt{{x}+\mathrm{1}}\:{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{2}} ^{\mathrm{6}} \sqrt{{x}−\mathrm{1}}\:{dx} \\ $$$${x}+\mathrm{1}={t}_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:{x}−\mathrm{1}={t}_{\mathrm{2}} ^{\mathrm{2}} \:\:\:\:\:\:{so}\:\:{dx}=\mathrm{2}{t}_{\mathrm{2}} {dt}_{\mathrm{2}} \\ $$$${dx}=\mathrm{2}{t}_{\mathrm{1}} {dt}_{\mathrm{1}} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\sqrt{\mathrm{3}}\:} ^{\sqrt{\mathrm{7}}\:} {t}_{\mathrm{1}} ×\mathrm{2}{t}_{\mathrm{1}} {dt}_{\mathrm{1}} \:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\sqrt{\mathrm{5}}} {t}_{\mathrm{2}} ×\mathrm{2}{t}_{\mathrm{2}} {dt}_{\mathrm{2}} \\ $$$${nw}\:{simple}\:{to}\:{get}\:{result}... \\ $$

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