Question Number 18650 by thukada last updated on 26/Jul/17
$$\int{dx}/{x}\sqrt{{x}^{\mathrm{4}} −\mathrm{1}} \\ $$
Answered by sma3l2996 last updated on 26/Jul/17
$${A}=\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}} \\ $$$${let}\:\:{t}=\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}\Rightarrow{dt}=\frac{\mathrm{2}{x}^{\mathrm{3}} }{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}{dx} \\ $$$$\frac{{dx}}{\:\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}=\frac{{dt}}{\mathrm{2}{x}^{\mathrm{3}} }\Leftrightarrow\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}=\frac{{dt}}{\mathrm{2}{x}^{\mathrm{4}} }=\frac{{dt}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left({t}\right)+{C} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}\right)+{C} \\ $$