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Question Number 26397 by abdo imad last updated on 25/Dec/17

find ∫ (dx/(x(√(1+x^2 )))) and calculate ∫_1 ^3 (dx/(x(√(1+x^2 ))))

$${find}\:\int\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{and}\:{calculate}\:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\frac{{dx}}{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$

Commented by abdo imad last updated on 26/Dec/17

let use the changement x=tanθ ∫ (dx/(x (√(1+x^2 ))))= ∫ (((1+tan^2 θ)dθ)/(tanθ (√(1+tan^2 ))))dθ= ∫ ((√(1+tan^2 θ))/(tanθ))dθ = ∫ (1/(cosθ tanθ))dθ= ∫ (dθ/(sinθ )) and by thechangement tan(θ/2)=t I= ∫ ((2dt)/(1+t^2 ))(((2t)/(1+t^2 )))^(−1) = ∫dt/t=ln/t/=ln/tan(((arctanx)/2) )/ ln ⇒ ∫ (dx/(x(√(1+x^2 ))))=ln/tan(((arctanx)/2) )/ +k ∫_1 ^3 (dx/(x(√(1+x^2 )))) = ln/tan(2^(−1) arctan3) /−ln/tan(2^(−1) arctan1)/ ==ln/tan(((arctan3)/2))/ −ln/tan((π/8))/ .

$${let}\:{use}\:{the}\:{changement}\:{x}={tan}\theta \\ $$$$\int\:\:\frac{{dx}}{{x}\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}=\:\:\:\:\int\:\:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta}{{tan}\theta\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} }}{d}\theta=\:\int\:\frac{\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}}{{tan}\theta}{d}\theta \\ $$$$=\:\:\int\:\:\:\:\:\frac{\mathrm{1}}{{cos}\theta\:{tan}\theta}{d}\theta=\:\:\int\:\:\frac{{d}\theta}{{sin}\theta\:}\:\:\:{and}\:{by}\:{thechangement}\:{tan}\left(\theta/\mathrm{2}\right)={t} \\ $$$${I}=\:\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{−\mathrm{1}} \:\:=\:\int{dt}/{t}={ln}/{t}/={ln}/{tan}\left(\frac{{arctanx}}{\mathrm{2}}\:\right)/\:{ln} \\ $$$$\Rightarrow\:\:\:\int\:\:\frac{{dx}}{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}={ln}/{tan}\left(\frac{{arctanx}}{\mathrm{2}}\:\right)/\:+{k} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \:\frac{{dx}}{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:=\:{ln}/{tan}\left(\mathrm{2}^{−\mathrm{1}} {arctan}\mathrm{3}\right)\:/−{ln}/{tan}\left(\mathrm{2}^{−\mathrm{1}} {arctan}\mathrm{1}\right)/ \\ $$$$=={ln}/{tan}\left(\frac{{arctan}\mathrm{3}}{\mathrm{2}}\right)/\:\:−{ln}/{tan}\left(\frac{\pi}{\mathrm{8}}\right)/\:. \\ $$

Answered by $@ty@m last updated on 25/Dec/17

x=tan y dx=sec^2 ydy I=∫((sec^2 ydy)/(tan y.sec y)) =∫cosec ydy =ln ∣cosec y−cot y∣+C =ln ∣cosec (tan^(−1) x)−cot (tan^(−1) x)∣+C

$${x}=\mathrm{tan}\:{y} \\ $$$${dx}=\mathrm{sec}\:^{\mathrm{2}} {ydy} \\ $$$${I}=\int\frac{\mathrm{sec}\:^{\mathrm{2}} {ydy}}{\mathrm{tan}\:{y}.\mathrm{sec}\:{y}} \\ $$$$=\int\mathrm{cosec}\:{ydy} \\ $$$$=\mathrm{ln}\:\mid\mathrm{cosec}\:{y}−\mathrm{cot}\:{y}\mid+{C} \\ $$$$=\mathrm{ln}\:\mid\mathrm{cosec}\:\left(\mathrm{tan}^{−\mathrm{1}} {x}\right)−\mathrm{cot}\:\left(\mathrm{tan}^{−\mathrm{1}} {x}\right)\mid+{C} \\ $$

Answered by $@ty@m last updated on 25/Dec/17

Another method: Let x=(1/t) dx=−(1/t^2 )dt ∫((−(1/t^2 )dt)/((1/t)(√(1+(1/t^2 ))))) ∫((−(1/t^2 )dt)/((1/t^2 )(√(t^2 +1)))) =∫((−dt)/(√(1+t^2 ))) =−ln ∣t+(√(1+t^2 ))∣+C =−ln ∣(1/x)+(√(1+(1/x^2 )))∣+C =−ln ∣((1+(√(1+x^2 )))/x)∣+C =ln ∣(x/(1+(√(1+x^2 ))))∣+C

$${Another}\:{method}: \\ $$$${Let}\:{x}=\frac{\mathrm{1}}{{t}} \\ $$$${dx}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt}}{\frac{\mathrm{1}}{{t}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}} \\ $$$$\int\frac{−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt}}{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$=\int\frac{−{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$=−\mathrm{ln}\:\mid{t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\mid+{C} \\ $$$$=−\mathrm{ln}\:\mid\frac{\mathrm{1}}{{x}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\mid+{C} \\ $$$$=−\mathrm{ln}\:\mid\frac{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\mid+{C} \\ $$$$=\mathrm{ln}\:\mid\frac{{x}}{\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\mid+{C} \\ $$

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