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Question Number 144042 by maryxxxx last updated on 20/Jun/21

Answered by mindispower last updated on 20/Jun/21

$$\int \frac{dx}{x^3\sqrt{1-{\left(\frac{10}{x}\right)}^2}}$$$$y=\frac{100}{x^2},y<1\Rightarrow dy=\frac{-200}{x^3}dx$$$$\Rightarrow \frac{-1}{200}\int \frac{dy}{\sqrt{1-y}}dy=\frac{1}{100}\sqrt{1-y}+c=\frac{\sqrt{1-\frac{100}{x^2}}}{100}+c$$$$$$$$$$

Answered by liberty last updated on 21/Jun/21

$$\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^3\sqrt{1-{100\mathrm{x}}^{-2}}}$$$$\mathrm{let}\mathrm{u}=\sqrt{1-{100\mathrm{x}}^{-2}}\wedge {\mathrm{u}}^2=1-{100\mathrm{x}}^{-2}$$$$\Rightarrow 2\mathrm{u}\mathrm{du}={200\mathrm{x}}^{-3}\mathrm{dx};\frac{\mathrm{dx}}{{\mathrm{x}}^3}=\frac{\mathrm{u}\mathrm{du}}{100}$$$$\mathrm{I}=\int \frac{1}{\mathrm{u}}\left(\frac{\mathrm{u}\mathrm{du}}{100}\right)=\frac{\mathrm{u}}{100}+\mathrm{c}$$$$=\frac{\sqrt{1-{100\mathrm{x}}^{-2}}}{100}+\mathrm{c}$$$$=\frac{\sqrt{\mathrm{x}-100}}{100\mathrm{x}}+\mathrm{c}$$

Answered by mathmax by abdo last updated on 21/Jun/21

$$\mathrm{\Psi }=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^2-100}}\Rightarrow \mathrm{\Psi }{=}_{\mathrm{x}=10\mathrm{cht}}\int \frac{10\mathrm{sht}\mathrm{dt}}{{100\mathrm{ch}}^2\mathrm{t}\times 10\mathrm{sht}}$$$$=\frac{1}{100}\int \frac{\mathrm{dt}}{{\mathrm{ch}}^2\mathrm{t}}=\frac{1}{50}\int \frac{\mathrm{dt}}{\mathrm{ch}\left(2\mathrm{t}\right)+1}{=}_{2\mathrm{t}=\mathrm{u}}\frac{1}{50}\int \frac{\mathrm{du}}{2(\mathrm{chu}+1)}$$$$=\frac{1}{100}\int \frac{\mathrm{du}}{\frac{{\mathrm{e}}^{\mathrm{u}}+{\mathrm{e}}^{-\mathrm{u}}}{2}+1}=\frac{1}{50}\int \frac{\mathrm{du}}{{\mathrm{e}}^{\mathrm{u}}+{\mathrm{e}}^{-\mathrm{u}}+2}$$$${=}_{{\mathrm{e}}^{\mathrm{u}}=\mathrm{z}}\frac{1}{50}\int \frac{\mathrm{dz}}{\mathrm{z}(\mathrm{z}+{\mathrm{z}}^{-1}+2)}=\frac{1}{50}\int \frac{\mathrm{dz}}{{\mathrm{z}}^2+2\mathrm{z}+1}$$$$=\frac{1}{50}\int \frac{\mathrm{dz}}{{(\mathrm{z}+1)}^2}=-\frac{1}{50(\mathrm{z}+1)}+\mathrm{K}$$$$=-\frac{1}{50({\mathrm{e}}^{\mathrm{u}}+1)}+\mathrm{K}=-\frac{1}{50(1+{\mathrm{e}}^{\frac{\mathrm{t}}{2}})}+\mathrm{K}$$$$\mathrm{t}=\mathrm{argch}\left(\frac{\mathrm{x}}{10}\right)=\log (\frac{\mathrm{x}}{10}+\sqrt{\frac{{\mathrm{x}}^2}{100}-1})\Rightarrow$$$$\mathrm{\Psi }=-\frac{1}{50(1+{\mathrm{e}}^{\frac{1}{2}\log (\frac{\mathrm{x}}{10}+\sqrt{\frac{{\mathrm{x}}^2}{100}-1})}}+\mathrm{K}$$$$=-\frac{1}{50(1+\sqrt{\frac{\mathrm{x}}{10}+\sqrt{\frac{{\mathrm{x}}^2}{100}-1})}}+\mathrm{K}(\mathrm{x}>10)$$

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