Integration Questions

Question Number 10147 by ridwan balatif last updated on 27/Jan/17

$$\int\sqrt{\frac{\mathrm{p}}{\mathrm{p}−\mathrm{1}}}\mathrm{dp}=...? \\$$

Commented by prakash jain last updated on 27/Jan/17

$${p}={x}^{\mathrm{2}} \\$$$${dp}=\mathrm{2}{xdx} \\$$$$=\int\sqrt{\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{2}{xdx} \\$$$$=\int\:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:{dx} \\$$$$\mathrm{substitute}\:{x}=\mathrm{cosh}\:{t} \\$$$${dx}=\mathrm{sinh}\:{t}\:{dt} \\$$$$=\int\mathrm{2cosh}^{\mathrm{2}} {t}\:{dt} \\$$$$=\int\left(\mathrm{1}+\mathrm{cosh}\:\mathrm{2}{t}\right){dt} \\$$$$=\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sinh}\:\mathrm{2}{t}\right)+{C} \\$$$$=\left({t}+\mathrm{sinh}\:{t}\mathrm{cosh}\:{t}\right)+{C} \\$$$$=\left(\mathrm{cosh}^{−\mathrm{1}} {x}+{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)+{C} \\$$$$=\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)+{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+{C} \\$$$$=\mathrm{ln}\:\left(\sqrt{{p}}+\sqrt{{p}−\mathrm{1}}\right)+\sqrt{{p}\left({p}−\mathrm{1}\right)}+{C} \\$$