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Question Number 137448 by liberty last updated on 03/Apr/21

$${\int }_0^{3/4}\frac{dx}{(x+1)\sqrt{x^2+1}}?$$

Answered by MJS_new last updated on 03/Apr/21

$$\int \frac{dx}{(x+1)\sqrt{x^2+1}}=$$$$[t=x+\sqrt{x^2+1}\iff x=\frac{t^2-1}{2t}\to dx=\frac{\sqrt{x^2+1}}{x+\sqrt{x^2+1}}dt]$$$$=2\int \frac{dt}{t^2+2t-1}=-\frac{\sqrt{2}}{2}\ln \frac{t+1+\sqrt{2}}{t+1-\sqrt{2}}=$$$$=-\frac{\sqrt{2}}{2}\ln \frac{-x+1+\sqrt{2(x^2+1)}}{x+1}$$$$\Rightarrow \mathrm{answer}\mathrm{is}(\ln (1+2\sqrt{2})-\frac{1}{2}\ln 7)\sqrt{2}$$

Commented by liberty last updated on 03/Apr/21

Commented by liberty last updated on 03/Apr/21

$$butigotintrigonometry$$

Answered by EDWIN88 last updated on 03/Apr/21

$$\mathrm{E}={\int }_0^{3/4}\frac{\mathrm{dx}}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+1}}$$$$\mathrm{let}\mathrm{x}=\tan \mathrm{t}\Rightarrow \mathrm{dx}={\sec }^2\mathrm{t}\mathrm{dt}$$$$\mathrm{E}={\int }^{}\frac{\sec {}^2\mathrm{t}}{(\tan \mathrm{t}+1)\sqrt{1+\tan {}^2\mathrm{t}}}\mathrm{dt}$$$$\mathrm{E}=\int \frac{\mathrm{dt}}{\sqrt{2}\sec (\mathrm{t}-\frac{\pi }{4})}=\frac{1}{\sqrt{2}}\ln \mid \sec (\mathrm{t}-\frac{\pi }{4})+\tan (\mathrm{t}-\frac{\pi }{4})\mid +\mathrm{C}$$$$=\frac{1}{\sqrt{2}}\ln \mid \frac{\sqrt{2}\sec \mathrm{t}+\tan \mathrm{t}-1}{1+\tan \mathrm{t}}\mid +\mathrm{C}$$$$=\frac{1}{\sqrt{2}}\ln \mid \frac{\sqrt{2+{2\mathrm{x}}^2}+\mathrm{x}-1}{1+\mathrm{x}}\mid +\mathrm{C}$$$$\mathrm{Now}{\int }_0^{3/4}\frac{dx}{(x+1)\sqrt{1+x^2}}$$$$=\frac{1}{\sqrt{2}}[\ln (2\sqrt{14}-1)-\ln \left(7\right)-\ln (\sqrt{2}-1)]$$$$\approx 0.569006$$

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