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Question Number 121601 by benjo_mathlover last updated on 09/Nov/20

$$\int \frac{\mathrm{dx}}{\mathrm{x}\sqrt{3+{\mathrm{x}}^2}}?$$

Answered by liberty last updated on 10/Nov/20

$$\mathrm{let}\mathrm{x}=\sqrt{3}\tan \gamma$$$$\mathrm{I}=\int \frac{\sqrt{3}\sec {}^2\gamma }{\sqrt{3}\tan \gamma \sqrt{3+3\tan {}^2\gamma }}\mathrm{d}\gamma$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\int \frac{\sec \gamma }{\tan \gamma }\mathrm{d}\gamma =\frac{1}{\sqrt{3}}\int \frac{1}{\sin \gamma }\mathrm{d}\gamma$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\int \mathrm{cosec}\gamma \mathrm{d}\gamma =\frac{1}{\sqrt{3}}\ln \mid \mathrm{cosec}\gamma -\cot \gamma \mid +\mathrm{c}$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\ln \mid \frac{\sqrt{3+{\mathrm{x}}^2}-\sqrt{3}}{\mathrm{x}}\mid +\mathrm{c}$$

Answered by Olaf last updated on 10/Nov/20

$$x=\sqrt{3}\sinh u$$$$\mathrm{I}=\int \frac{\sqrt{3}\cosh u}{\sqrt{3}\sinh u\sqrt{3+{3\sinh }^{2}u}}du$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\int \frac{du}{\sinh u}$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\int \frac{du}{\frac{e^u-e^{-u}}{2}}$$$$\mathrm{I}=\frac{2}{\sqrt{3}}\int \frac{e^{u}du}{e^{2u}-1}$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\int [\frac{1}{e^u-1}-\frac{1}{e^u+1}]e^{u}du$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\ln \mid \frac{e^u-1}{e^u+1}\mid$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\ln \mid \frac{e^{u/2}-e^{-u/2}}{e^{u/2}+e^{-u/2}}\mid$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\ln \mid \tanh \left(\frac{u}{2}\right)\mid$$$$\mathrm{I}=\frac{1}{\sqrt{3}}\ln \mid \tanh \left(\frac{1}{2}\mathrm{argsinh}\left(\frac{x}{\sqrt{3}}\right)\right)\mid$$

Answered by Dwaipayan Shikari last updated on 10/Nov/20

$$\int \frac{dx}{x\sqrt{3+x^2}}$$$$\int \frac{{sec}^2\theta }{\sqrt{3}tan\theta sec\theta }d\theta x=\sqrt{3}tan\theta$$$$=\frac{1}{\sqrt{3}}\int \frac{1}{sin\theta }d\theta =\frac{1}{3}log\left(tan\frac{\theta }{2}\right)+C$$$$=\frac{1}{\sqrt{3}}log\left(\frac{-\sqrt{3}+\sqrt{3+x^2}}{x}\right)+C$$

Answered by TANMAY PANACEA last updated on 10/Nov/20

$$\int \frac{xdx}{x^2\sqrt{3+x^2}}$$$$t^2=3+x^2\to 2tdt=xdx$$$$\int \frac{tdt}{(t^2-3)\times t}=\int \frac{dt}{t^2-3}$$$$=\frac{1}{2\sqrt{3}}ln\left(\frac{t-\sqrt{3}}{t+\sqrt{3}}\right)+c$$$$=\frac{1}{2\sqrt{3}}ln\left(\frac{\sqrt{3+x^2}-\sqrt{3}}{\sqrt{3+x^2}+\sqrt{3}}\right)+c$$

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