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Question Number 89714 by abdomathmax last updated on 18/Apr/20

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

Commented by mathmax by abdo last updated on 19/Apr/20

$$parametricmethodletf\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{t+\sqrt{2+x^2}}witht>0$$$$f^{{}^{\prime }}\left(t\right)=-{\int }_0^{\mathrm{\infty }}\frac{dx}{{(t+\sqrt{2+x^2})}^2}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{dx}{{(t+\sqrt{2+x^2})}^2}=-f^{{}^{\prime }}\left(t\right)$$$$changementx=\sqrt{2}sh\left(u\right)give$$$$f\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{\sqrt{2}ch\left(u\right)du}{t+\sqrt{2}ch\left(u\right)}=\sqrt{2}{\int }_0^{\mathrm{\infty }}\frac{\frac{e^u+e^{-u}}{2}}{t+\sqrt{2}\times \frac{e^u+e^{-u}}{2}}du$$$$=\sqrt{2}{\int }_0^{\mathrm{\infty }}\frac{e^u+e^{-u}}{2t+\sqrt{2}{e}^u+\sqrt{2}{e}^{-u}}du{=}_{e^u=z}\sqrt{2}{\int }_1^{\mathrm{\infty }}\frac{z+z^{-1}}{2t+\sqrt{2}z+\sqrt{2}{z}^{-1}}\times \frac{dz}{z}$$$$=\sqrt{2}{\int }_1^{+\mathrm{\infty }}\frac{z+z^{-1}}{2tz+\sqrt{2}{z}^2+\sqrt{2}}dz$$$$=\sqrt{2}{\int }_1^{+\mathrm{\infty }}\frac{z^2+1}{z(\sqrt{2}{z}^2+2tz+\sqrt{2})}dzdecopositionofg\left(z\right)=\frac{z^2+1}{z(\sqrt{2}{z}^2+2tz+\sqrt{2})}$$$$\sqrt{2}{z}^2+2tz+\sqrt{2}=0\to {\mathrm{\Delta }}^{{}^{\prime }}=t^2-2<0ift<\sqrt{2}$$$$z_1=\frac{-t+i\sqrt{2-t^2}}{\sqrt{2}}and{z}_2=\frac{-t-i\sqrt{2-t^2}}{\sqrt{2}}$$$$g\left(z\right)=\frac{z^2+1}{\sqrt{2}z(z-z_1)(z-z_2)}=\frac{a}{z}+\frac{b}{z-z_1}+\frac{c}{z-z_2}$$$$a=\frac{1}{\sqrt{2}{z}_1{z}_2}=\frac{1}{\sqrt{2}},b=\frac{z_1^2+1}{\sqrt{2}{z}_1\sqrt{2}i\sqrt{2-t^2}}=\frac{1+z_1^2}{2{iz}_1\sqrt{2-t^2}}$$$$b=\frac{z_2^2+1}{\sqrt{2}{z}_2(-\sqrt{2}i\sqrt{2-t^2})}=\frac{1+z_2^2}{-2{iz}_2\sqrt{2-t^2}}\Rightarrow$$$${\int }_1^{+\mathrm{\infty }}g\left(z\right)dz={\int }_1^{+\mathrm{\infty }}(\frac{a}{z}+\frac{b}{z-z_1}+\frac{c}{z-z_2})dz$$$$={[aln\mid z\mid +bln\mid z-z_1\mid +cln\mid z-z_2\mid ]}_1^{+\mathrm{\infty }}resttosimplifybandc$$$$becontinued...$$

Answered by MJS last updated on 18/Apr/20

$$\int \frac{dx}{{(1+\sqrt{x^2+2})}^2}=$$$$[t=\frac{x+\sqrt{x^2+2}}{\sqrt{2}}\to dx=\frac{\sqrt{2(x^2+2)}}{x+\sqrt{x^2+2}}dt]$$$$=\sqrt{2}\int \frac{t^2+1}{{(t^2+\sqrt{2}t+1)}^2}dt=$$$$\left[\mathrm{Ostrogradski}\right]$$$$=\frac{\sqrt{2}t+2}{t^2+\sqrt{2}t+1}+2\sqrt{2}\int \frac{dt}{t^2+\sqrt{2}t+1}=$$$$=\frac{\sqrt{2}t+2}{t^2+\sqrt{2}t+1}+4\arctan (\sqrt{2}t+1)=$$$$=\frac{x^2+x+1-x\sqrt{x^2+2}}{x^2+1}+4\arctan (x+1+\sqrt{x^2+2})+C$$$$\Rightarrow$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{(1+\sqrt{x^2+2})}^2}=\frac{\pi }{2}-1$$

Commented by turbo msup by abdo last updated on 18/Apr/20

$$thankyousir.$$

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