Question Number 136883 by bramlexs22 last updated on 27/Mar/21
$$\mathrm{Given}\:\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{7}^{{a}} −\mathrm{2}\right)=\:\mathrm{log}\:_{\mathrm{7}} \left(\mathrm{5}^{{a}} +\mathrm{2}\right) \\ $$$$.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\int_{{a}} ^{\mathrm{e}} \:\frac{\mathrm{1}+\mathrm{ln}\:\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{x}} +\mathrm{x}^{−\mathrm{x}} }\:\mathrm{dx}\:. \\ $$
Answered by Olaf last updated on 27/Mar/21
![log_5 (7^a −2) = log_7 (5^a +2) ⇒ a = 1 (trivial solution) Ω = ∫_1 ^e ((1+lnx)/(x^x +x^(−x) )) dx Ω = ∫_1 ^e ((1+lnx)/(e^(xlnx) +e^(−xlnx) )) dx Ω = ∫_1 ^e ((d(xlnx))/(e^(xlnx) +e^(−xlnx) )) Ω = ∫_1 ^e (e^(xlnx) /(1+e^(2xlnx) )) d(xlnx) Ω = [arctan(e^(xlnx) )]_1 ^e Ω = arctan(e^e )−(π/4)](Q136890.png)
$$\mathrm{log}_{\mathrm{5}} \left(\mathrm{7}^{{a}} −\mathrm{2}\right)\:=\:\mathrm{log}_{\mathrm{7}} \left(\mathrm{5}^{{a}} +\mathrm{2}\right) \\ $$$$\Rightarrow\:{a}\:=\:\mathrm{1}\:\left(\mathrm{trivial}\:\mathrm{solution}\right) \\ $$$$\Omega\:=\:\int_{\mathrm{1}} ^{{e}} \frac{\mathrm{1}+\mathrm{ln}{x}}{{x}^{{x}} +{x}^{−{x}} }\:{dx} \\ $$$$\Omega\:=\:\int_{\mathrm{1}} ^{{e}} \frac{\mathrm{1}+\mathrm{ln}{x}}{{e}^{{x}\mathrm{ln}{x}} +{e}^{−{x}\mathrm{ln}{x}} }\:{dx} \\ $$$$\Omega\:=\:\int_{\mathrm{1}} ^{{e}} \frac{{d}\left({x}\mathrm{ln}{x}\right)}{{e}^{{x}\mathrm{ln}{x}} +{e}^{−{x}\mathrm{ln}{x}} }\: \\ $$$$\Omega\:=\:\int_{\mathrm{1}} ^{{e}} \frac{{e}^{{x}\mathrm{ln}{x}} }{\mathrm{1}+{e}^{\mathrm{2}{x}\mathrm{ln}{x}} }\:{d}\left({x}\mathrm{ln}{x}\right) \\ $$$$\Omega\:=\:\left[\mathrm{arctan}\left({e}^{{x}\mathrm{ln}{x}} \right)\right]_{\mathrm{1}} ^{{e}} \\ $$$$\Omega\:=\:\mathrm{arctan}\left({e}^{{e}} \right)−\frac{\pi}{\mathrm{4}} \\ $$
Answered by liberty last updated on 27/Mar/21
