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Question Number 184514 by BOYQOBILOV last updated on 08/Jan/23

Answered by SEKRET last updated on 19/Jan/23

$$\mathbf{t}=\mathbf{x}+\sqrt{{\mathbf{x}}^2+1}{/}_0^1\to {/}_1^{1+\sqrt{2}}$$$$\mathbf{dt}=1+\frac{\mathbf{x}}{\sqrt{{\mathbf{x}}^2+1}}\mathbf{dx}=\frac{\mathbf{x}+\sqrt{{\mathbf{x}}^2+1}}{\sqrt{{\mathbf{x}}^2+1}}\mathbf{dx}$$$${\int }_1^{1+\sqrt{2}}\frac{\sqrt{\mathbf{t}}}{\sqrt{{\mathbf{x}}^2+1}}\cdot \frac{\sqrt{{\mathbf{x}}^2+1}}{\mathbf{t}}\mathbf{dt}=$$$${\int }_1^{1+\sqrt{2}}\frac{1}{\sqrt{\mathbf{t}}}\mathbf{dt}=2\sqrt{\mathbf{t}}{\mid }_1^{1+\sqrt{2}}=2\cdot (\sqrt{1+\sqrt{2}}-1)$$$$\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{L}\mathit{A}\mathit{Z}\mathit{I}\mathit{Z}\mathit{A}\mathit{B}\mathit{D}\mathit{U}\mathit{V}\mathit{A}\mathit{L}\mathit{I}\mathit{Y}\mathit{E}\mathit{V}$$

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