Question Number 99114 by M±th+et+s last updated on 18/Jun/20
$${calculate}: \\ $$$$\int\sqrt{{x}}{sinh}^{−\mathrm{1}} \left({x}\right){dx} \\ $$$${where}\:{sinh}^{−\mathrm{1}} \left({x}\right)\:{is}\:{the}\:{inverse}\:{hyperbolic}\: \\ $$$${sine}\:{function} \\ $$$$ \\ $$$$ \\ $$
Answered by Rio Michael last updated on 19/Jun/20
![By parts let u = sinh^(−1) x and v′ = (√x) ⇒ u′ = (1/( (√(1 + x^2 )))) and v = ((2x^(3/2) )/3) ⇒∫ (√x) sinh^(−1) xdx = [((2x^(3/2) )/3) sinh^(−1) x]−(2/3)∫(x^(3/2) /( (√(x^2 + 1)))) dx to be continued.....](https://www.tinkutara.com/question/Q99177.png)
$$\:\mathrm{By}\:\mathrm{parts}\: \\ $$$$\mathrm{let}\:{u}\:=\:\mathrm{sinh}^{−\mathrm{1}} {x}\:\mathrm{and}\:{v}'\:=\:\sqrt{{x}} \\ $$$$\Rightarrow\:{u}'\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:+\:{x}^{\mathrm{2}} }}\:\mathrm{and}\:{v}\:=\:\frac{\mathrm{2}{x}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}} \\ $$$$\Rightarrow\int\:\sqrt{{x}}\:\mathrm{sinh}\:^{−\mathrm{1}} {xdx}\:=\:\left[\frac{\mathrm{2}{x}^{\mathrm{3}/\mathrm{2}} }{\mathrm{3}}\:\mathrm{sinh}^{−\mathrm{1}} {x}\right]−\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{x}^{\mathrm{3}/\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{continued}….. \\ $$