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

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

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

$$\sqrt{1+x^2}=t\Rightarrow \frac{x}{\sqrt{1+x^2}}=\frac{dt}{dx}$$$$={\int }_1^{\mathrm{\infty }}\frac{1}{x^2}dt={\int }_1^{\mathrm{\infty }}\frac{1}{t^2-1}={\left[\frac{1}{2}log\left(\frac{t-1}{t+1}\right)\right]}_1^{\mathrm{\infty }}={doesn}^{\prime }tconverge$$

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