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Question Number 124532 by liberty last updated on 04/Dec/20

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

Answered by bemath last updated on 04/Dec/20

$$\vartheta \left(x\right)=\int \frac{dx}{x^3\sqrt{1-x^{-2}}}=\int \frac{x^{-3}}{\sqrt{1-x^{-2}}}dx$$$$lettingz=1-x^{-2}\to dz=2x^{-3}dx$$$$\vartheta \left(x\right)=\frac{1}{2}\int \frac{dz}{\sqrt{z}}=\frac{1}{2}\int z^{-\frac{1}{2}}dz$$$$\vartheta \left(x\right)=\sqrt{z}+c=\sqrt{1-\frac{1}{x^2}}+c$$$$\vartheta \left(x\right)=\frac{\sqrt{x^2-1}}{x}+c$$

Answered by MJS_new last updated on 04/Dec/20

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{absolutely}\mathrm{crazy}\mathrm{and}\mathrm{I}\mathrm{should}\mathrm{be}\mathrm{kept}$$$$\mathrm{away}\mathrm{from}\mathrm{students}...$$$$\mathrm{this}\mathrm{is}\mathrm{no}\mathrm{fun}\mathrm{to}\mathrm{transform}\mathrm{but}\mathrm{I}\mathrm{had}\mathrm{lots}\mathrm{of}$$$$\mathrm{fun}\mathrm{transforming}\mathrm{it}\mathrm{and}\mathrm{the}\mathrm{integral}...\mathrm{look}$$$$\mathrm{at}\mathrm{the}\mathrm{integral}...{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{it}supereasy?!$$$$t=\sqrt{2x(x+\sqrt{x^2-1})}\iff x=\frac{t^2}{2\sqrt{t^2-1}}$$$$\to dx=\frac{t(t^2-2)}{2{(t^2-1)}^{3/2}}dt$$$$\int \frac{dx}{x^2\sqrt{x^2-1}}=8\int \frac{{(t^2-1)}^{3/2}}{t^4(t^2-2)}dx=$$$$=\mathrm{♡}4\int \frac{dt}{t^3}\mathrm{♡}=-\frac{2}{t^2}=-\frac{1}{x(x+\sqrt{x^2-1})}=$$$$=\frac{\sqrt{x^2-1}-x}{x}=\frac{\sqrt{x^2-1}}{x}-1=\frac{\sqrt{x^2-1}}{x}+C$$

Commented by bemath last updated on 04/Dec/20

$$hahaha..youarenotcrazyprof.but$$$$youareamasterofmathand$$$$santuy.wkwkwk...$$

Answered by Ar Brandon last updated on 04/Dec/20

$$\mathcal{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^2-1}}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^3\sqrt{1-\frac{1}{{\mathrm{x}}^2}}}$$$$1-\frac{1}{{\mathrm{x}}^2}=\mathrm{u}\Rightarrow \frac{2}{{\mathrm{x}}^3}\mathrm{dx}=\mathrm{du}$$$$\mathcal{I}=\int \frac{1}{\sqrt{\mathrm{u}}}\cdot \frac{\mathrm{du}}{2}=\sqrt{\mathrm{u}}+\mathcal{C}=\sqrt{1-\frac{1}{{\mathrm{x}}^2}}+\mathcal{C}$$

Answered by malwan last updated on 04/Dec/20

$$x=secy\Rightarrow dx=secytanydy$$$$\therefore \int \frac{secytany}{{sec}^{2}y\sqrt{{sec}^{2}y-1}}dy=\int \frac{tany}{secy\sqrt{{tan}^{2}y}}dy$$$$=\int \frac{dy}{secy}=\int cosydy=\frac{\sqrt{x^2-1}}{x}+C$$

Answered by mathmax by abdo last updated on 04/Dec/20

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

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