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Question Number 37961 by prof Abdo imad last updated on 19/Jun/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{dx}{x^2+\sqrt{1+x^2}}.$$

Answered by MJS last updated on 20/Jun/18

$$\int \frac{dx}{x^2+\sqrt{x^2+1}}=\int \frac{x^2}{x^4-x^2-1}dx-\int \frac{\sqrt{x^2+1}}{x^4-x^2-1}dx=$$$$$$$$\int \frac{x^2}{x^4-x^2-1}dx=$$$$={\mathcal{N}}_1(\int \frac{dx}{x-\frac{\sqrt{2}}{2}\sqrt{1+\sqrt{5}}}-\int \frac{dx}{x+\frac{\sqrt{2}}{2}\sqrt{1+\sqrt{5}}})+{\mathcal{N}}_2\int \frac{dx}{x^2-\frac{1}{2}+\frac{\sqrt{5}}{2}}=$$$$=\frac{\sqrt{10}}{20}\sqrt{1+\sqrt{5}}\ln \mid \frac{x-\frac{\sqrt{2}}{2}\sqrt{1+\sqrt{5}}}{x+\frac{\sqrt{2}}{2}\sqrt{1+\sqrt{5}}}\mid +\frac{\sqrt{10}}{10}\sqrt{-1+\sqrt{5}}\arctan \left(\frac{\sqrt{2}}{2}\sqrt{1+\sqrt{5}}x\right)$$$$$$$$\int \frac{\sqrt{x^2+1}}{x^4-x^2-1}dx=$$$$[t=\frac{x}{\sqrt{x^2+1}}\to dx=\sqrt{{(x^2+1)}^3}dt]$$$$\int \frac{dt}{t^4+t^2-1}=$$$$={\mathcal{N}}_3(\int \frac{dt}{t-\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}}-\int \frac{dt}{t+\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}})+{\mathcal{N}}_4\int \frac{dt}{t^2+\frac{1}{2}+\frac{\sqrt{5}}{2}}=$$$$=\frac{\sqrt{10}}{20}\sqrt{1+\sqrt{5}}\ln \mid \frac{t-\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}}{t+\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}}\mid -\frac{\sqrt{10}}{10}\sqrt{-1+\sqrt{5}}\arctan \left(\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}t\right)=$$$$=\frac{\sqrt{10}}{20}\sqrt{1+\sqrt{5}}\ln \mid \frac{\frac{x}{\sqrt{x^2+1}}-\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}}{\frac{x}{\sqrt{x^2+1}}+\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}}\mid -\frac{\sqrt{10}}{10}\sqrt{-1+\sqrt{5}}\arctan \left(\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}\frac{x}{\sqrt{x^2+1}}\right)$$$$$$$$=\frac{\sqrt{10}}{20}(\sqrt{1+\sqrt{5}}(\ln \mid \frac{x-\frac{\sqrt{2}}{2}\sqrt{1+\sqrt{5}}}{x+\frac{\sqrt{2}}{2}\sqrt{1+\sqrt{5}}}\mid -\ln \mid \frac{\frac{x}{\sqrt{x^2+1}}-\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}}{\frac{x}{\sqrt{x^2+1}}+\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}}\mid )+2\sqrt{-1+\sqrt{5}}(\arctan \left(\frac{\sqrt{2}}{2}\sqrt{1+\sqrt{5}}x\right)+\arctan \left(\frac{\sqrt{2}}{2}\sqrt{-1+\sqrt{5}}\frac{x}{\sqrt{x^2+1}}\right)))+C$$$$$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{x^2+\sqrt{x^2+1}}=\frac{\sqrt{10}}{20}(\sqrt{1+\sqrt{5}}\ln (2+\sqrt{5}+2\sqrt{2+\sqrt{5}})+\sqrt{-1+\sqrt{5}}(\pi +2\arcsin \left(\frac{-1+\sqrt{5}}{2}\right)))\approx 1.39021$$

Commented by math khazana by abdo last updated on 20/Jun/18

$$thankyousirforthishardwork.$$

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