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Question Number 136313 by aurpeyz last updated on 20/Mar/21

$$\int \frac{1}{x^2\sqrt{x^2-9}}dx$$

Answered by liberty last updated on 20/Mar/21

$$\int \frac{dx}{x^3\sqrt{1-9x^{-2}}}=\int \frac{x^{-3}}{\sqrt{1-9x^{-2}}}dx$$$$let{w}^2=1-9x^{-2}\Rightarrow 2wdw=18x^{-3}dx$$$$I=\frac{1}{9}\int \frac{wdw}{\sqrt{w^2}}=\frac{1}{9}\int dw$$$$I=\frac{1}{9}w+c=\frac{1}{9}\sqrt{\frac{x^2-9}{x^2}}+c$$$$$$

Answered by Olaf last updated on 20/Mar/21

$$\mathrm{F}\left(x\right)=\int \frac{dx}{x^2\sqrt{x^2-9}}$$$$\mathrm{Let}x=3\mathrm{ch}u$$$$\mathrm{F}\left(x\right)=\int \frac{3\mathrm{sh}u}{{9\mathrm{ch}}^{2}u\sqrt{{9\mathrm{ch}}^{2}u-9}}du$$$$\mathrm{F}\left(x\right)=\frac{1}{3}\int \frac{\mathrm{sh}u}{{\mathrm{ch}}^{2}u\times 3\mathrm{sh}u}du$$$$\mathrm{F}\left(x\right)=\frac{1}{9}\int \frac{du}{{\mathrm{ch}}^{2}u}$$$$\mathrm{F}\left(x\right)=\frac{1}{9}\int (1-{\mathrm{th}}^{2}u)du$$$$\mathrm{F}\left(x\right)=\frac{1}{9}\mathrm{th}u+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{1}{9}\mathrm{th}\left(\mathrm{argch}\frac{x}{3}\right)+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{1}{9}.\frac{e^{\mathrm{argch}\frac{x}{3}}-e^{-\mathrm{argch}\frac{x}{3}}}{e^{\mathrm{argch}\frac{x}{3}}+e^{-\mathrm{argch}\frac{x}{3}}}+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{1}{9}.\frac{e^{\ln (\frac{x}{3}+\sqrt{\frac{x^2}{9}-1})}-e^{-\ln (\frac{x}{3}+\sqrt{\frac{x^2}{9}-1})}}{e^{\ln (\frac{x}{3}+\sqrt{\frac{x^2}{9}-1})}+e^{-\ln (\frac{x}{3}+\sqrt{\frac{x^2}{9}-1})}}+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{1}{9}.\frac{\frac{x}{3}+\sqrt{\frac{x^2}{9}-1}-\frac{1}{\frac{x}{3}+\sqrt{\frac{x^2}{9}-1}}}{\frac{x}{3}+\sqrt{\frac{x^2}{9}-1}+\frac{1}{\frac{x}{3}+\sqrt{\frac{x^2}{9}-1}}}+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{1}{9}.\frac{\frac{x}{3}+\sqrt{\frac{x^2}{9}-1}-(\frac{x}{3}-\sqrt{\frac{x^2}{9}-1})}{\frac{x}{3}+\sqrt{\frac{x^2}{9}-1}+(\frac{x}{3}-\sqrt{\frac{x^2}{9}-1})}+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{1}{9}.\frac{2\sqrt{\frac{x^2}{9}-1}}{\frac{2x}{3}}+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{1}{3}.\frac{\sqrt{\frac{x^2}{9}-1}}{x}+\mathrm{C}$$$$\mathrm{F}\left(x\right)=\frac{1}{9x}\sqrt{x^2-9}+\mathrm{C}$$

Answered by mathmax by abdo last updated on 20/Mar/21

$$\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^2-9}}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=3\mathrm{cht}\Rightarrow$$$$\mathrm{I}=\int \frac{3\mathrm{sht}}{{9\mathrm{ch}}^2\mathrm{t}.3\mathrm{sht}}\mathrm{dt}=\frac{2}{9}\int \frac{\mathrm{dt}}{1+\mathrm{ch}\left(2\mathrm{t}\right)}{=}_{2\mathrm{t}=\mathrm{u}}\frac{2}{9}\int \frac{\mathrm{du}}{2(1+\mathrm{chu})}$$$$=\frac{1}{9}\int \frac{\mathrm{du}}{1+\frac{{\mathrm{e}}^{\mathrm{u}}+{\mathrm{e}}^{-\mathrm{u}}}{2}}=\frac{2}{9}\int \frac{\mathrm{du}}{2+{\mathrm{e}}^{\mathrm{u}}+{\mathrm{e}}^{-\mathrm{u}}}{=}_{{\mathrm{e}}^{\mathrm{u}}=\mathrm{y}}\frac{2}{9}\int \frac{\mathrm{dy}}{\mathrm{y}(2+\mathrm{y}+{\mathrm{y}}^{-1})}$$$$=\frac{2}{9}\int \frac{\mathrm{dy}}{2\mathrm{y}+{\mathrm{y}}^2+1}=\frac{2}{9}\int \frac{\mathrm{dy}}{{(\mathrm{y}+1)}^2}=-\frac{2}{9(\mathrm{y}+1)}+\mathrm{C}$$$$\mathrm{y}={\mathrm{e}}^{\mathrm{u}}={\mathrm{e}}^{2\mathrm{t}}\mathrm{and}\mathrm{t}=\mathrm{argch}\left(\frac{\mathrm{x}}{3}\right)=\ln (\frac{\mathrm{x}}{3}+\sqrt{\frac{{\mathrm{x}}^2}{9}-1})\Rightarrow$$$$\mathrm{y}={(\frac{\mathrm{x}}{3}+\sqrt{\frac{{\mathrm{x}}^2}{9}-1})}^2\Rightarrow \mathrm{I}=-\frac{2}{9({(\frac{\mathrm{x}}{3}+\sqrt{\frac{{\mathrm{x}}^2}{9}-1})}^2+1)}+\mathrm{C}$$$$$$

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