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Question Number 120654 by peter frank last updated on 01/Nov/20

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{[\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}]}^{\mathrm{n}}}$$

Answered by mindispower last updated on 01/Nov/20

$$x=sh\left(t\right),n>1$$$$\iff {\int }_0^{\mathrm{\infty }}\frac{ch\left(t\right)dt}{{(ch\left(t\right)+sh\left(t\right))}^n}$$$$={\int }_0^{\mathrm{\infty }}\frac{e^t+e^{-t}}{2e^{nt}}dt$$$$=\frac{1}{2}{[\frac{e^{(1-n)t}}{1-n}-\frac{e^{-(1+n)t}}{1+n}]}_0^{\mathrm{\infty }}$$$$=\frac{1}{2}[\frac{1}{n-1}+\frac{1}{n+1}]=\frac{n}{n^2-1}$$

Commented by peter frank last updated on 01/Nov/20

$$\mathrm{thank}\mathrm{you}$$

Commented by mindispower last updated on 03/Nov/20

$$withepleasur$$

Answered by MJS_new last updated on 01/Nov/20

$$\int \frac{dx}{{(x+\sqrt{x^2+1})}^n}=$$$$[t=x+\sqrt{x^2+1}\to dx=\frac{\sqrt{x^2+1}}{x+\sqrt{x^2+1}}dt]$$$$=\frac{1}{2}\int \frac{t^2+1}{t^{n+2}}dt=\frac{1}{2}\int \frac{1}{t^{n+2}}+\frac{1}{t^n}dt=$$$$=-\frac{1}{2(n+1)t^{n+1}}-\frac{1}{2(n-1)t^{n-1}}=[\Rightarrow ]$$$$=-\frac{(n+1)t^2-(n-1)}{2(n^2-1)t^{n+1}}=$$$$=\frac{(x+n\sqrt{x^2+1}){(-x+\sqrt{x^2+1})}^n}{1-n^2}+C$$$$[\Rightarrow \underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{(x+\sqrt{x^2+1})}^n}=$$$$={[-\frac{1}{2(n+1)t^{n+1}}-\frac{1}{2(n-1)t^{n-1}}]}_1^{\mathrm{\infty }}=$$$$=\frac{n}{1-n^2}\mathrm{with}n\in \mathbb{Z}\mathrm{\setminus }\{\pm 1\}$$

Commented by peter frank last updated on 01/Nov/20

$$\mathrm{thank}\mathrm{you}$$

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