Question Number 212467 by Hanuda354 last updated on 14/Oct/24
$$\mathrm{Use}\:\:\mathrm{the}\:\:\mathrm{integration}\:\:\mathrm{by}\:\:\mathrm{parts}. \\ $$$$ \\ $$$$\int\:\frac{\mathrm{2}}{\:\sqrt{{r}^{\mathrm{2}} −\mathrm{4}}}\:{dr}\: \\ $$
Answered by mehdee7396 last updated on 14/Oct/24
$${r}=\mathrm{2}{coshx}\Rightarrow{dr}=\mathrm{2}{sinhxdx} \\ $$$$\Rightarrow{I}=\int\frac{\mathrm{4}{sinhxdx}}{\mathrm{2}\sqrt{{cosh}^{\mathrm{2}} {x}−\mathrm{1}}}=\mathrm{2}\int{dx}=\mathrm{2}{x}+{c} \\ $$$$=\mathrm{2}{coh}^{−\mathrm{1}} \frac{{r}}{\mathrm{2}}+{c}=\mathrm{2}{ln}\left({r}+\sqrt{{r}^{\mathrm{2}} −\mathrm{4}}\right)+{c}'\:\checkmark \\ $$$$ \\ $$
Commented by Frix last updated on 14/Oct/24
$$\mathrm{Yes}\:\mathrm{but}\:\mathrm{it}\:\mathrm{says}\:``\mathrm{use}\:\mathrm{the}\:\mathrm{integration}\:\mathrm{by}\:\mathrm{parts}'' \\ $$