Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 183806 by cortano1 last updated on 30/Dec/22

∫ ((√(x+(√(x^2 +25))))/x) dx =?

$$\:\:\int\:\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}}{{x}}\:{dx}\:=? \\ $$

Answered by Ar Brandon last updated on 30/Dec/22

I=∫((√(x+(√(x^2 +25))))/x)dx , x=5shϑ =∫((√(5shϑ+5chϑ))/(5shϑ))(5chϑdϑ)=(√5)∫(((e^(2ϑ) +1)/(e^(2ϑ) −1)))e^(ϑ/2) dϑ, ϑ=2φ =2(√5)∫(((e^(4φ) +1)/(e^(4φ) −1)))e^φ dφ=2(√5)∫(((t^4 +1)/(t^4 −1)))dt=2(√5)∫(1+(2/(t^4 −1)))dt, t=e^φ =2(√5)∫(1+(1/(t^2 −1))−(1/(t^2 +1)))dt=2(√5)(t+(1/2)ln∣((t−1)/(t+1))∣−arctan(t))+C =2(√5)(e^(ϑ/2) +(1/2)ln∣e^(ϑ/2) −1∣−(1/2)ln∣e^(ϑ/2) +1∣−arctan(e^(ϑ/2) ))+C =2(√5)(√(x+(√(x^2 +25))))+(√5)ln∣(((√(x+(√(x^2 +25))))−1)/( (√(x+(√(x^2 +25))))+1))∣−2(√5)arctan((√(x+(√(x^2 +25)))))+C

$${I}=\int\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}}{{x}}{dx}\:,\:{x}=\mathrm{5sh}\vartheta \\ $$$$\:\:=\int\frac{\sqrt{\mathrm{5sh}\vartheta+\mathrm{5ch}\vartheta}}{\mathrm{5sh}\vartheta}\left(\mathrm{5ch}\vartheta{d}\vartheta\right)=\sqrt{\mathrm{5}}\int\left(\frac{{e}^{\mathrm{2}\vartheta} +\mathrm{1}}{{e}^{\mathrm{2}\vartheta} −\mathrm{1}}\right){e}^{\frac{\vartheta}{\mathrm{2}}} {d}\vartheta,\:\vartheta=\mathrm{2}\phi \\ $$$$\:\:=\mathrm{2}\sqrt{\mathrm{5}}\int\left(\frac{{e}^{\mathrm{4}\phi} +\mathrm{1}}{{e}^{\mathrm{4}\phi} −\mathrm{1}}\right){e}^{\phi} {d}\phi=\mathrm{2}\sqrt{\mathrm{5}}\int\left(\frac{{t}^{\mathrm{4}} +\mathrm{1}}{{t}^{\mathrm{4}} −\mathrm{1}}\right){dt}=\mathrm{2}\sqrt{\mathrm{5}}\int\left(\mathrm{1}+\frac{\mathrm{2}}{{t}^{\mathrm{4}} −\mathrm{1}}\right){dt},\:{t}={e}^{\phi} \\ $$$$\:\:=\mathrm{2}\sqrt{\mathrm{5}}\int\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt}=\mathrm{2}\sqrt{\mathrm{5}}\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid−\mathrm{arctan}\left({t}\right)\right)+{C} \\ $$$$\:\:=\mathrm{2}\sqrt{\mathrm{5}}\left({e}^{\frac{\vartheta}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid{e}^{\frac{\vartheta}{\mathrm{2}}} −\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid{e}^{\frac{\vartheta}{\mathrm{2}}} +\mathrm{1}\mid−\mathrm{arctan}\left({e}^{\frac{\vartheta}{\mathrm{2}}} \right)\right)+{C} \\ $$$$\:\:=\mathrm{2}\sqrt{\mathrm{5}}\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}+\sqrt{\mathrm{5}}\mathrm{ln}\mid\frac{\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}−\mathrm{1}}{\:\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}+\mathrm{1}}\mid−\mathrm{2}\sqrt{\mathrm{5}}\mathrm{arctan}\left(\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{25}}}\right)+{C} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com