0-1-1-1-4x-dx-

Question Number 186263 by mokys last updated on 02/Feb/23

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}}}\:{dx} \\ $$$$ \\ $$

Answered by cortano1 last updated on 03/Feb/23

let (1/(2(√x))) = tan t ⇒2(√x) = cot t ⇒4x = cot^2 t ⇒4dx = 2cot t (−csc^2 t) dt I=∫ (1/( (√(1+tan^2 t)))) (−2cot t csc^2 t)dt I= −2∫ cos t ((cos t)/(sin t)) . (1/(sin^2 t)) dt I=−2∫ ((1−sin^2 t)/(sin^3 t)) dt I=−2∫(csc^3 t−csc t) dt

$$\:{let}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\:=\:\mathrm{tan}\:{t}\:\Rightarrow\mathrm{2}\sqrt{{x}}\:=\:\mathrm{cot}\:{t} \\ $$$$\Rightarrow\mathrm{4}{x}\:=\:\mathrm{cot}\:^{\mathrm{2}} {t}\: \\ $$$$\Rightarrow\mathrm{4}{dx}\:=\:\mathrm{2cot}\:{t}\:\left(−\mathrm{csc}^{\mathrm{2}} \:{t}\right)\:{dt} \\ $$$${I}=\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {t}}}\:\left(−\mathrm{2cot}\:{t}\:\mathrm{csc}^{\mathrm{2}} {t}\right){dt} \\ $$$${I}=\:−\mathrm{2}\int\:\mathrm{cos}\:{t}\:\frac{\mathrm{cos}\:{t}}{\mathrm{sin}\:{t}}\:.\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} {t}}\:{dt}\: \\ $$$${I}=−\mathrm{2}\int\:\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {t}}{\mathrm{sin}\:^{\mathrm{3}} {t}}\:{dt} \\ $$$${I}=−\mathrm{2}\int\left(\mathrm{csc}^{\mathrm{3}} {t}−\mathrm{csc}\:{t}\right)\:{dt}\: \\ $$$$ \\ $$

Answered by Mathspace last updated on 03/Feb/23

let (1/(4x))=t ⇒x=(1/(4t)) ⇒ I=−∫_(1/4) ^(+∞) (√(1+t))(((−dt)/(4t^2 ))) 4I=∫_(1/4) ^(+∞) ((√(1+t))/t^2 )dt (1+t=z^2 ) =∫_((√5)/2) ^∞ (z/((z^2 −1)^2 ))(2z)dz =∫_((√5)/2) z(((2z)/((z^2 −1)^2 )))dz (by parts) =[−(z/(z^2 −1))]_((√5)/2) ^∞ −∫_((√5)/2) ^∞ (−(1/(z^2 −1)))dz =((√5)/(2((5/4)−1)))+(1/2)∫_((√5)/2) ^∞ ((1/(z−1))−(1/(z+1)))dz =2(√5) +(1/2)[ln∣((z−1)/(z+1))∣]_((√5)/2) ^∞ =2(√5)−(1/2)ln∣((((√5)/2)−1)/(((√5)/2)+1))∣ =2(√5)−(1/2)ln((((√5)−2)/( (√5)+2)))

$${let}\:\frac{\mathrm{1}}{\mathrm{4}{x}}={t}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{4}{t}}\:\Rightarrow \\ $$$${I}=−\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{+\infty} \sqrt{\mathrm{1}+{t}}\left(\frac{−{dt}}{\mathrm{4}{t}^{\mathrm{2}} }\right) \\ $$$$\mathrm{4}{I}=\int_{\frac{\mathrm{1}}{\mathrm{4}}} ^{+\infty} \frac{\sqrt{\mathrm{1}+{t}}}{{t}^{\mathrm{2}} }{dt}\:\:\:\left(\mathrm{1}+{t}={z}^{\mathrm{2}} \right) \\ $$$$=\int_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} \frac{{z}}{\left({z}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\left(\mathrm{2}{z}\right){dz} \\ $$$$=\int_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} \:\:{z}\left(\frac{\mathrm{2}{z}}{\left({z}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\right){dz}\:\:\left({by}\:{parts}\right) \\ $$$$=\left[−\frac{{z}}{{z}^{\mathrm{2}} −\mathrm{1}}\right]_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} −\int_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} \left(−\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{1}}\right){dz} \\ $$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} \left(\frac{\mathrm{1}}{{z}−\mathrm{1}}−\frac{\mathrm{1}}{{z}+\mathrm{1}}\right){dz} \\ $$$$=\mathrm{2}\sqrt{\mathrm{5}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\mid\right]_{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\infty} \\ $$$$=\mathrm{2}\sqrt{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{1}}{\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}+\mathrm{1}}\mid \\ $$$$=\mathrm{2}\sqrt{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{2}}{\:\sqrt{\mathrm{5}}+\mathrm{2}}\right) \\ $$

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