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Question Number 122748 by bemath last updated on 19/Nov/20

$$\underset{0}{\stackrel{3}{\int }}\frac{dx}{(3-x)\sqrt{x^2+1}}?$$

Answered by MJS_new last updated on 19/Nov/20

$$\int \frac{dx}{(3-x)\sqrt{x^2+1}}=$$$$[t=x+\sqrt{x^2+1}\to dx=\frac{\sqrt{x^2+1}}{x+\sqrt{x^2+1}}]$$$$=-2\int \frac{dt}{t^2-6t-1}=-2\int \frac{dt}{(t-3+\sqrt{10})(t-3+\sqrt{10})}=$$$$=\frac{\sqrt{10}}{10}\int (\frac{1}{t-3+\sqrt{10}}-\frac{1}{t-3-\sqrt{10}})dt=$$$$=\frac{\sqrt{10}}{10}\ln \frac{t-3+\sqrt{10}}{t-3-\sqrt{10}}=$$$$=\frac{\sqrt{10}}{10}\ln \mid \frac{3x+1+\sqrt{10(x^2+1)}}{3-x}\mid +C$$$$\underset{0}{\stackrel{3}{\int }}...{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{converge}$$

Answered by liberty last updated on 19/Nov/20

$$since\underset{x\to 3^{-}}{\lim }(3-x).\frac{1}{(3-x)\sqrt{x^2+1}}=\frac{1}{\sqrt{10}}$$$$itfollowsthat\underset{0}{\stackrel{3}{\int }}\frac{dx}{(3-x)\sqrt{x^2+1}}diverges$$

Answered by Bird last updated on 19/Nov/20

$$A={\int }_0^3\frac{dx}{(3-x)\sqrt{x^2+1}}wedothe$$$$changementx=sht\Rightarrow$$$$A={\int }_0^{argsh\left(3\right)}\frac{cht}{(3-sht)cht}dt$$$$={\int }_0^{ln(3+\sqrt{10})}\frac{dt}{3-\frac{e^t-e^{-t}}{2}}$$$$={\int }_0^{ln(3+\sqrt{10})}\frac{2dt}{6-e^t+e^{-t}}$$$${=}_{e^t=x}{\int }_1^{3+\sqrt{10}}\frac{2dx}{x(6-x+x^{-1})}$$$$=2{\int }_1^{3+\sqrt{10}}\frac{dx}{6x-x^2+1}$$$$=-2{\int }_1^{3+\sqrt{10}}\frac{dx}{x^2-6x-1}$$$${\mathrm{\Delta }}^{{}^{\prime }}=9+1=10\Rightarrow x_1=3+\sqrt{10}$$$$x_2=3-\sqrt{10}\Rightarrow$$$$I=-2{\int }_1^{3+\sqrt{10}}\frac{dx}{(x-3-\sqrt{10})(x-3+\sqrt{10})}$$$$=-\frac{2}{2\sqrt{10}}{\int }_1^{3+\sqrt{10}}(\frac{1}{x-3-\sqrt{10}}-\frac{1}{x-3+\sqrt{10}})dx$$$$=-\frac{1}{\sqrt{10}}{[ln\mid \frac{x-3-\sqrt{10}}{x-3+\sqrt{10}}\mid ]}_1^{3+\sqrt{10}}=-\mathrm{\infty }$$$$thisintegraldkverges..\sqrt{!}$$

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