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Question-200960




Question Number 200960 by sonukgindia last updated on 27/Nov/23
Answered by Frix last updated on 27/Nov/23
t=1+(√(x^2 +1))  ⇒  ∫...=∫(1−(1/t))dt=t−ln t =  =1+(√(x^2 +1))−ln (1+(√(x^2 +1))) +C  [1+(√(x^2 +1))−ln (1+(√(x^2 +1)))]_(−a) ^a =0
$${t}=\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow \\ $$$$\int…=\int\left(\mathrm{1}−\frac{\mathrm{1}}{{t}}\right){dt}={t}−\mathrm{ln}\:{t}\:= \\ $$$$=\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{ln}\:\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\:+{C} \\ $$$$\left[\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{ln}\:\left(\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\right]_{−{a}} ^{{a}} =\mathrm{0} \\ $$
Answered by Frix last updated on 27/Nov/23
=∫_(−1) ^1 (((√(x^2 +1))−1)/x)dx  f(x)=(((√(x^2 +1))−1)/x)  f(−x)=−f(x)
$$=\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}}{{x}}{dx} \\ $$$${f}\left({x}\right)=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}}{{x}} \\ $$$${f}\left(−{x}\right)=−{f}\left({x}\right) \\ $$
Answered by Sutrisno last updated on 29/Nov/23
∫_(−1) ^1 (((√(x^2 +1))+(x−1))/( (√(x^2 +1))+x+1)).(((√(x^2 +1))−(x+1))/( (√(x^2 +1))−(x+1)))dx  ∫_(−1) ^1 (((√(x^2 +1))−1))/( x))dx  let: (√(x^2 +1))=t→x=(√(t^2 −1))→dx=(t/x)dt  ∫((t−1)/( (√(t^2 −1)))).(t/x)dt  ∫((t−1)/( (√(t^2 −1)))).(t/( (√(t^2 −1))))dt  ∫((t(t−1))/( (t+1)(t−1)))dt  ∫(t/( (t+1)))dt  =t−ln(t+1)  =(√(x^2 +1))−ln((√(x^2 +1))+1)∣_(−1) ^1   =(√2)−ln((√2)+1)−((√2)−ln((√2)+1))  =0
$$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\left({x}−\mathrm{1}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+{x}+\mathrm{1}}.\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\left({x}+\mathrm{1}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\left({x}+\mathrm{1}\right)}{dx} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\left.\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{1}\right)}{\:{x}}{dx} \\ $$$${let}:\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}={t}\rightarrow{x}=\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\rightarrow{dx}=\frac{{t}}{{x}}{dt} \\ $$$$\int\frac{{t}−\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}.\frac{{t}}{{x}}{dt} \\ $$$$\int\frac{{t}−\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}.\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}{dt} \\ $$$$\int\frac{{t}\left({t}−\mathrm{1}\right)}{\:\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right)}{dt} \\ $$$$\int\frac{{t}}{\:\left({t}+\mathrm{1}\right)}{dt} \\ $$$$={t}−{ln}\left({t}+\mathrm{1}\right) \\ $$$$=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{ln}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}\right)\mid_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$=\sqrt{\mathrm{2}}−{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)−\left(\sqrt{\mathrm{2}}−{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\right) \\ $$$$=\mathrm{0} \\ $$$$ \\ $$

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