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Question Number 184552 by Frix last updated on 08/Jan/23

$$I_1=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}(\frac{\sqrt{x+\sqrt{x^2+1}}}{\sqrt{x^2+1}}-\frac{\sqrt{2}}{\sqrt{x}})dx=?$$$$I_2=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}(\frac{\sqrt{x^2+1}}{\sqrt{x+\sqrt{x^2+1}}}-\frac{\sqrt{x}}{\sqrt{2}})dx=?$$

Answered by SEKRET last updated on 08/Jan/23

$$\mathbf{t}=\mathbf{x}+\sqrt{{\mathbf{x}}^2+1}$$$$$$

Commented by Frix last updated on 08/Jan/23

$$\mathrm{Yes}.\mathrm{But}\mathrm{please}\mathrm{find}\mathrm{the}\mathrm{values}!$$

Answered by SEKRET last updated on 08/Jan/23

$$\int \sqrt{\frac{\mathbf{x}+\sqrt{{\mathbf{x}}^2+1}}{{\mathbf{x}}^2+1}}\mathbf{dx}$$$$\mathbf{u}=\mathbf{x}+\sqrt{{\mathbf{x}}^2+1}\mathbf{du}=\frac{\mathbf{x}+\sqrt{{\mathbf{x}}^2+1}}{\sqrt{{\mathbf{x}}^2+1}}\mathbf{dx}$$$$\int \frac{1}{\sqrt{\mathbf{u}}}\mathbf{du}=2\sqrt{\mathbf{u}}+\mathbf{c}=2\sqrt{\mathbf{x}+\sqrt{{\mathbf{x}}^2+1}}+\mathbf{C}$$$$\int \sqrt{\frac{{\mathbf{x}}^2+1}{\sqrt{{\mathbf{x}}^2+1}+\mathbf{x}}}\mathbf{dx}$$$$\mathbf{u}=\sqrt{{\mathbf{x}}^2+1}+\mathbf{x}$$$$\int \frac{{({\mathbf{u}}^2+1)}^2}{{\mathbf{u}}^{\frac{7}{2}}}\mathbf{du}=\int {\mathbf{u}}^{\frac{1}{2}}+2{\mathbf{u}}^{\frac{-1}{2}}+{\mathbf{u}}^{\frac{-7}{2}}\mathbf{du}$$$$=\frac{2}{3}\cdot {\mathbf{u}}^{\frac{3}{2}}+4\cdot {\mathbf{u}}^{\frac{1}{2}}-\frac{2}{5}\cdot {\mathbf{u}}^{-\frac{5}{2}}+\mathbf{C}$$$$$$

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