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Question Number 109403 by abony1303 last updated on 23/Aug/20

$$\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}=\ln \mid \mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\mid +C\mathrm{Proof}?$$

Answered by bobhans last updated on 23/Aug/20

$$letx=a\tan s\Rightarrow dx=a\sec {}^{2}sds$$$$I=\int \frac{a\sec {}^{2}sds}{\sqrt{a^2(\tan {}^{2}s+1)}}=\int \sec sds$$$$I=\ln \mid \sec s+\tan s\mid +c$$$$=\ln \mid \frac{x}{a}+\frac{\sqrt{x^2+a^2}}{a}\mid +c$$$$=\ln \mid x+\sqrt{x^2+a^2}\mid -\ln a+c$$$$=\ln \mid x+\sqrt{x^2+a^2}\mid +C;C=-\ln a+c$$

Answered by 1549442205PVT last updated on 23/Aug/20

$$\mathrm{Since}\mid \mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\mid =\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}(\mathrm{due}$$$$\mathrm{to}\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}>0\mathrm{\forall }\mathrm{x}\in \mathrm{R}),\mathrm{so}\mathrm{we}\mathrm{have}$$$$\frac{\mathrm{d}}{\mathrm{dx}}(\ln \mid \mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\mid +C)=\frac{\mathrm{d}}{\mathrm{dx}}\left[\ln (\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2})\right]$$$$=\frac{1}{\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\times \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2})=$$$$=\frac{1}{\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\times (1+\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}})$$$$=\frac{1}{\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}\times \frac{\mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}=\frac{1}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}$$$$\mathrm{This}\mathrm{shows}\mathrm{that}$$$$\int \frac{\mathrm{dx}}{\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}=\ln \mid \mathrm{x}+\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}\mid +\mathrm{C}(\mathbf{Q}.\mathbf{E}.\mathbf{D})$$

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