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Question Number 212467 by Hanuda354 last updated on 14/Oct/24

$$\mathrm{Use}\mathrm{the}\mathrm{integration}\mathrm{by}\mathrm{parts}.$$$$$$$$\int \frac{2}{\sqrt{r^2-4}}dr$$

Answered by mehdee7396 last updated on 14/Oct/24

$$r=2coshx\Rightarrow dr=2sinhxdx$$$$\Rightarrow I=\int \frac{4sinhxdx}{2\sqrt{{cosh}^{2}x-1}}=2\int dx=2x+c$$$$=2{coh}^{-1}\frac{r}{2}+c=2ln(r+\sqrt{r^2-4})+c^{\prime }✓$$$$$$

Commented by Frix last updated on 14/Oct/24

$$\mathrm{Yes}\mathrm{but}\mathrm{it}\mathrm{says}‘‘\mathrm{use}\mathrm{the}\mathrm{integration}\mathrm{by}{\mathrm{parts}}^{″}$$

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