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Question Number 109472 by john santu last updated on 24/Aug/20

Answered by 1549442205PVT last updated on 24/Aug/20

$$\mathrm{Put}\mathrm{F}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{1+{\mathrm{x}}^2}}$$$$\mathrm{Putting}\frac{1}{{\mathrm{x}}^2}+1={\mathrm{u}}^2\Rightarrow 2\mathrm{udu}=\frac{-2}{{\mathrm{x}}^3}\mathrm{dx}$$$$\Rightarrow \mathrm{dx}=-{\mathrm{ux}}^3\mathrm{du},\sqrt{{\mathrm{x}}^2+1}=\mathrm{ux}$$$$\Rightarrow {\mathrm{x}}^2\sqrt{1+{\mathrm{x}}^2}={\mathrm{ux}}^3\Rightarrow \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{1+{\mathrm{x}}^2}}=-\mathrm{du}$$$$\mathrm{Therefore},\mathrm{F}=-\int \mathrm{du}=-\mathrm{u}+\mathrm{C}=-\frac{\sqrt{1+{\mathrm{x}}^2}}{\mathrm{x}}+\mathrm{C}$$

Answered by $@y@m last updated on 24/Aug/20

$$Letx=\tan \theta$$$$dx=\sec {}^2\theta d\theta$$$$\int \frac{\sec {}^2\theta d\theta }{\tan {}^2\theta \sec \theta }$$$$\int \frac{\sec \theta d\theta }{\tan {}^2\theta }$$$$\int \frac{\cos \theta }{\sin {}^2\theta }d\theta$$$$-\frac{1}{\sin \theta }+C$$$$-\mathrm{cosec}\theta +C$$$$-\sqrt{1+\cot {}^2\theta }+C$$$$-\sqrt{1+\frac{1}{x^2}}+C$$$$-\frac{\sqrt{1+x^2}}{x}+C$$

Answered by bobhans last updated on 24/Aug/20

$$letx=\frac{1}{v}\Rightarrow dx=-\frac{dv}{v^2}$$$$I=\int \frac{v^2}{\sqrt{1+\frac{1}{v^2}}}\times (-\frac{dv}{v^2})$$$$I=-\int \frac{v}{\sqrt{1+v^2}}dv=-\frac{1}{2}\int \frac{d(1+v^2)}{\sqrt{1+v^2}}$$$$I=-\frac{1}{2}.2\sqrt{1+v^2}+c=-\sqrt{1+\frac{1}{x^2}}+c$$$$\frac{\mathcal{B}obhans}{⩽\bullet ⩾}$$

Answered by mathmax by abdo last updated on 24/Aug/20

$$\mathrm{A}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\sqrt{1+{\mathrm{x}}^2}}\Rightarrow \mathrm{A}{=}_{\mathrm{x}=\mathrm{sht}}\int \frac{\mathrm{cht}\mathrm{dt}}{{\mathrm{sh}}^2\mathrm{t}\mathrm{cht}}=\int \frac{2\mathrm{dt}}{\mathrm{ch}\left(2\mathrm{t}\right)-1}$$$$=2\int \frac{\mathrm{dt}}{\frac{{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{e}}^{-2\mathrm{t}}}{2}-1}=4\int \frac{\mathrm{dt}}{{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{e}}^{-2\mathrm{t}}-2}{=}_{{\mathrm{e}}^{2\mathrm{t}}=\mathrm{z}}4\int \frac{\mathrm{dz}}{2\mathrm{z}(\mathrm{z}+{\mathrm{z}}^{-1}-2)}$$$$=2\int \frac{\mathrm{dz}}{{\mathrm{z}}^2+1-2\mathrm{z}}=2\int \frac{\mathrm{dz}}{{(\mathrm{z}-1)}^2}=\frac{-2}{\mathrm{z}-1}+\mathrm{C}=\frac{2}{1-\mathrm{z}}+\mathrm{C}$$$$=\frac{2}{1-{\mathrm{e}}^{2\mathrm{t}}}+\mathrm{C}=\frac{2}{1-{(\mathrm{x}+\sqrt{1+\mathrm{x}})}^2}+\mathrm{C}(\mathrm{t}=\mathrm{argshx}=\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})$$$$$$

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