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Question Number 175467 by manish54 last updated on 31/Aug/22

Commented by Frix last updated on 01/Sep/22

$$\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{(x^2+1)}^{1/2}}={\left[\sinh x\right]}_1^{\mathrm{\infty }}\mathrm{does}\mathrm{not}\mathrm{converge}$$$$\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{(x^2+1)}^{3/2}}={\left[\frac{x}{{(x^2+1)}^{1/2}}\right]}_1^{\mathrm{\infty }}=1-\frac{1}{\sqrt{2}}$$

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