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Question Number 124217 by bramlexs22 last updated on 01/Dec/20

∫_0 ^∞  (dx/(x (√(1+x^2 )))) ?

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{{dx}}{{x}\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:? \\ $$

Answered by Dwaipayan Shikari last updated on 01/Dec/20

(√(1+x^2 )) =t⇒(x/( (√(1+x^2 ))))=(dt/dx)  =∫_1 ^∞ (1/x^2 )dt=∫_1 ^∞ (1/(t^2 −1))=[(1/2)log(((t−1)/(t+1)))]_1 ^∞ =doesn′t converge

$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:={t}\Rightarrow\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}=\frac{{dt}}{{dx}} \\ $$$$=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }{dt}=\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}=\left[\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\right)\right]_{\mathrm{1}} ^{\infty} ={doesn}'{t}\:{converge} \\ $$

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