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Question Number 212048 by Spillover last updated on 28/Sep/24

Answered by Ghisom last updated on 28/Sep/24

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{\sqrt{x}\sqrt{x^2+1}\sqrt{x+\sqrt{x^2+1}}}=$$$$[t=x+\sqrt{x^2+1}]$$$$=\sqrt{2}\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{dt}{t\sqrt{t^2-1}}=$$$$[u=t+\sqrt{t^2-1}]$$$$=2\sqrt{2}\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\frac{du}{u^2+1}={\left[2\sqrt{2}\arctan u\right]}_1^{\mathrm{\infty }}=$$$$=\frac{\sqrt{2}\pi }{2}$$

Commented by Spillover last updated on 28/Sep/24

$$great.thanks$$

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