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Question Number 37067 by math khazana by abdo last updated on 08/Jun/18
find ∫ (dx/((x+1)(√(1+x^2 ))))
$${find}\:\int\:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$
Commented by math khazana by abdo last updated on 09/Jun/18
changement x+1=(1/t) give t=(1/(x+1)) I = ∫ (t/( (√(1+((1/t)−1)^2 )))) ((−dt)/t^2 ) =−∫ (dt/(t(√(1+ (((t−1)^2 )/t^2 ))))) =−∫ ((t dt)/(t(√(t^2 +t^2 −2t +1)))) =−∫ (dt/( (√(2t^2 −2t +1))))?but 2t^2 −2t +1 =2{ t^2 −t +(1/2)} =2 { (t−(1/2))^2 +(1/2) −(1/4)}=2{ (t−(1/2))^2 +(1/4)} changemnt t−(1/2)=(1/2) u?give I = −(1/( (√2))) ∫ (1/((1/2)(√(1+u^2 )))) (du/2) =−(1/( (√2))) ∫ (du/( (√(1+u^2 )))) +c =−(1/( (√2))) ln(u +(√(1+u^2 )) )+c = −(1/( (√2)))ln( 2t−1 +(√(1+(2t−1)^2 ))) +c =−(1/( (√2)))ln{ (2/(x+1)) −1 +(√(1+((2/(x+1)) −1)^2 )) +c
$${changement}\:{x}+\mathrm{1}=\frac{\mathrm{1}}{{t}}\:{give}\:{t}=\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$ \\ $$$${I}\:=\:\int\:\:\:\:\:\frac{{t}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}}{{t}}−\mathrm{1}\right)^{\mathrm{2}} }}\:\frac{−{dt}}{{t}^{\mathrm{2}} } \\ $$$$=−\int\:\:\:\:\:\frac{{dt}}{{t}\sqrt{\mathrm{1}+\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} }}} \\ $$$$=−\int\:\:\:\:\:\frac{{t}\:{dt}}{{t}\sqrt{{t}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \:−\mathrm{2}{t}\:+\mathrm{1}}} \\ $$$$=−\int\:\:\:\:\:\:\frac{{dt}}{\:\sqrt{\mathrm{2}{t}^{\mathrm{2}} \:−\mathrm{2}{t}\:+\mathrm{1}}}?{but} \\ $$$$\mathrm{2}{t}^{\mathrm{2}} \:−\mathrm{2}{t}\:+\mathrm{1}\:=\mathrm{2}\left\{\:{t}^{\mathrm{2}} −{t}\:+\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$$$=\mathrm{2}\:\left\{\:\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{4}}\right\}=\mathrm{2}\left\{\:\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:\:+\frac{\mathrm{1}}{\mathrm{4}}\right\} \\ $$$${changemnt}\:{t}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\:{u}?{give} \\ $$$${I}\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{{du}}{\mathrm{2}} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\:\:\frac{{du}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}\:+{c} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{ln}\left({u}\:+\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\:\right)+{c} \\ $$$$=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\left(\:\mathrm{2}{t}−\mathrm{1}\:+\sqrt{\mathrm{1}+\left(\mathrm{2}{t}−\mathrm{1}\right)^{\mathrm{2}} }\right)\:+{c} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\left\{\:\frac{\mathrm{2}}{{x}+\mathrm{1}}\:−\mathrm{1}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}}{{x}+\mathrm{1}}\:−\mathrm{1}\right)^{\mathrm{2}} }\:+{c}\right. \\ $$
Answered by MJS last updated on 08/Jun/18
see my answer to qu. 37018 (1)
$$\mathrm{see}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{qu}.\:\mathrm{37018}\:\left(\mathrm{1}\right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18
t=(1/(x+1)) x+1=(1/t) dx=−(1/t^2 )dt =−∫(dt/t^2 )×(t/( (√(1+((1/t)−1)^2 )))) =−∫(dt/t)×(t/( (√(t^2 +(1−2t+t^2 ))))) =−∫(dt/( (√(2t^2 −2t+1)))) =−(1/( (√2)))∫(dt/( (√(t^2 −t+(1/2))))) =((−1)/( (√2)))∫(dt/( (√(t^2 −2.t.(1/2)+(1/4)+(1/2)−(1/4))))) =((−1)/( (√2)))∫(dt/( (√((t−(1/2))^2 +((1/( (√2))))^2 )))) now use formula ∫(dx/( (√(x^2 +a^2 ))))
$${t}=\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$${x}+\mathrm{1}=\frac{\mathrm{1}}{{t}}\:\:\:\:{dx}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$=−\int\frac{{dt}}{{t}^{\mathrm{2}} }×\frac{{t}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{1}}{{t}}−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$$=−\int\frac{{dt}}{{t}}×\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{2}{t}+{t}^{\mathrm{2}} \right)}} \\ $$$$=−\int\frac{{dt}}{\:\sqrt{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}}} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} −{t}+\frac{\mathrm{1}}{\mathrm{2}}}} \\ $$$$=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{2}.{t}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}}} \\ $$$$ \\ $$$$=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{dt}}{\:\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }} \\ $$$${now}\:{use}\:{formula}\:\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$
Commented by math khazana by abdo last updated on 09/Jun/18
sir Tanmay you have commited a error of calculus because t^2 −t +(1/2) =(t−(1/2))^2 +(1/4) !
$${sir}\:{Tanmay}\:{you}\:{have}\:{commited}\:{a}\:{error}\:{of} \\ $$$${calculus}\:{because}\:{t}^{\mathrm{2}} \:−{t}\:+\frac{\mathrm{1}}{\mathrm{2}}\:=\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\:! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jun/18
yes in place of(1/(2 )) i put 1...rectifying inred colour
$${yes}\:{in}\:{place}\:{of}\frac{\mathrm{1}}{\mathrm{2}\:\:}\:{i}\:{put}\:\mathrm{1}…{rectifying}\:{inred}\:{colour} \\ $$

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