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Question Number 87540 by Power last updated on 04/Apr/20

Commented by abdomathmax last updated on 05/Apr/20

$$letI={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x+\sqrt{1+x^2})}^2}changementx=sh\left(t\right)give$$$$I={\int }_0^{\mathrm{\infty }}\frac{ch\left(t\right)dt}{{(sh\left(t\right)+cht)}^2}={\int }_0^{\mathrm{\infty }}\frac{ch\left(t\right)}{e^{2t}}dt$$$$={\int }_0^{\mathrm{\infty }}{e}^{-2t}\left(\frac{e^t+e^{-t}}{2}\right)dt=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-t}dt+\frac{1}{2}{\int }_0^{\mathrm{\infty }}{e}^{-3t}dt$$$$=\frac{1}{2}{[-e^{-t}]}_0^{+\mathrm{\infty }}+\frac{1}{2}{[-\frac{1}{3}{e}^{-3t}]}_0^{+\mathrm{\infty }}$$$$=\frac{1}{2}\left(1\right)+\frac{1}{6}=\frac{3}{6}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$$

Answered by MJS last updated on 04/Apr/20

$$\int \frac{dx}{{(x+\sqrt{1+x^2})}^2}=\int (2x^2+1-2x\sqrt{x^2+1})dx=$$$$=\frac{2}{3}{x}^3+x-\frac{2}{3}\sqrt{{(x^2+1)}^3}+C$$$$\Rightarrow \underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{{(x+\sqrt{1+x^2})}^2}=\frac{2}{3}$$

Commented by Power last updated on 05/Apr/20

$$\mathrm{sir}\mathrm{please}\mathrm{solve}\mathrm{pls}$$

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