Question Number 195252 by Spillover last updated on 28/Jul/23
$$\int_{\boldsymbol{{spillover}}} \:\:\:\:\:\:\frac{{dx}}{\:\sqrt{{e}^{\mathrm{5}{x}} }\:\sqrt{\left({e}^{\mathrm{2}{x}} +{e}^{−\mathrm{2}{x}} \right)^{\mathrm{3}} }} \\ $$
Answered by Frix last updated on 28/Jul/23
$$=\int\frac{\mathrm{e}^{\frac{{x}}{\mathrm{2}}} }{\:\left(\mathrm{e}^{\mathrm{4}{x}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}= \\ $$$$=\frac{\mathrm{e}^{\frac{{x}}{\mathrm{2}}} }{\mathrm{2}\left(\mathrm{e}^{\mathrm{4}{x}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }+\frac{\mathrm{3}}{\mathrm{4}}\int\frac{\mathrm{e}^{\frac{{x}}{\mathrm{2}}} }{\left(\mathrm{e}^{\mathrm{4}{x}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dx}= \\ $$$$=\frac{\mathrm{e}^{\frac{{x}}{\mathrm{2}}} }{\mathrm{2}\left(\mathrm{e}^{\mathrm{4}{x}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }+\frac{\mathrm{3e}^{\frac{{x}}{\mathrm{2}}} }{\mathrm{2}}\:_{\mathrm{2}} \mathrm{F}_{\mathrm{1}} \:\left(\frac{\mathrm{1}}{\mathrm{8}},\:\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{9}}{\mathrm{8}};\:−\mathrm{e}^{\mathrm{4}{x}} \right)\:= \\ $$$$=\frac{\mathrm{e}^{\frac{{x}}{\mathrm{2}}} }{\mathrm{2}}\left(\left(\mathrm{e}^{\mathrm{4}{x}} +\mathrm{1}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{3}\:_{\mathrm{2}} \mathrm{F}_{\mathrm{1}} \:\left(\frac{\mathrm{1}}{\mathrm{8}},\:\frac{\mathrm{1}}{\mathrm{2}};\:\frac{\mathrm{9}}{\mathrm{8}};\:−\mathrm{e}^{\mathrm{4}{x}} \right)\right)+{C} \\ $$
Commented by Spillover last updated on 03/Aug/23
$${very}\:{nice} \\ $$