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If-4-e-x-6-e-x-9-e-x-4-e-x-dx-Ax-B-log-9e-2x-4-C-then-

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Question Number 63865 by mmkkmm000m last updated on 10/Jul/19
If ∫ ((4 e^x + 6 e^(−x) )/(9 e^x − 4 e^(−x) ))dx=Ax+B log(9e^(2x) −4)+C, then
$$\mathrm{If}\:\int\:\:\frac{\mathrm{4}\:{e}^{{x}} +\:\mathrm{6}\:{e}^{−{x}} }{\mathrm{9}\:{e}^{{x}} −\:\mathrm{4}\:{e}^{−{x}} }{dx}={Ax}+{B}\:\mathrm{log}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)+{C}, \\ $$$$\mathrm{then} \\ $$
Commented by kaivan.ahmadi last updated on 10/Jul/19
(Ax+Blog(9e^(2x) −4))^′ =((4e^x +6e^(−x) )/(9e^x −4e^(−x) )) but the left hand side is equal to A+B×((18e^(2x) )/(9e^(2x) −4))×(1/(ln10))=A+(B/(ln10))×((18e^x )/(9e^x −4e^(−x) ))= ((9Ae^x ln10−4Ae^(−x) ln10+18Be^x )/(ln10(9e^x −4e^(−x) )))= (((9A+((18)/(ln10))B)e^x −4Ae^(−x) )/(9e^x −4e^(−x) ))⇒ { ((4A=6⇒A=(3/2))),((9A+((18)/(ln10))B=4⇒((18)/(ln10))B=((−19)/2)⇒B=((−19ln10)/(36)))) :}
$$\left({Ax}+{Blog}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)\right)^{'} =\frac{\mathrm{4}{e}^{{x}} +\mathrm{6}{e}^{−{x}} }{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} }\:\:\:{but} \\ $$$${the}\:{left}\:{hand}\:{side}\:{is}\:{equal}\:{to} \\ $$$${A}+{B}×\frac{\mathrm{18}{e}^{\mathrm{2}{x}} }{\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}}×\frac{\mathrm{1}}{{ln}\mathrm{10}}={A}+\frac{{B}}{{ln}\mathrm{10}}×\frac{\mathrm{18}{e}^{{x}} }{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} }= \\ $$$$\frac{\mathrm{9}{Ae}^{{x}} {ln}\mathrm{10}−\mathrm{4}{Ae}^{−{x}} {ln}\mathrm{10}+\mathrm{18}{Be}^{{x}} }{{ln}\mathrm{10}\left(\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} \right)}= \\ $$$$\frac{\left(\mathrm{9}{A}+\frac{\mathrm{18}}{{ln}\mathrm{10}}{B}\right){e}^{{x}} −\mathrm{4}{Ae}^{−{x}} }{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} }\Rightarrow \\ $$$$\begin{cases}{\mathrm{4}{A}=\mathrm{6}\Rightarrow{A}=\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{9}{A}+\frac{\mathrm{18}}{{ln}\mathrm{10}}{B}=\mathrm{4}\Rightarrow\frac{\mathrm{18}}{{ln}\mathrm{10}}{B}=\frac{−\mathrm{19}}{\mathrm{2}}\Rightarrow{B}=\frac{−\mathrm{19}{ln}\mathrm{10}}{\mathrm{36}}}\end{cases} \\ $$
Commented by mathmax by abdo last updated on 10/Jul/19
let I =∫ ((4e^x +6e^(−x) )/(9e^x −4 e^(−x) ))dx[⇒I =∫ ((4e^(2x) +6)/(9e^(2x) −4))dx changement e^(2x) =t give I =∫ ((4t+6)/(9t−4)) (dt/(2t)) =∫ ((2t+3)/(t(9t−4)))dt let decompose F(t)=((2t+3)/(t(9t−4))) F(t)=(a/t) + (b/(9t−4)) a=lim_(t→0) tF(t)=−(3/4) b=lim_(t→(4/9)) (9t−4)F(t) =((2×(4/9)+3)/(4/9)) =((8+27)/4) =((35)/4) ⇒ F(t)=−(3/(4t)) +((35)/(4(9t−4))) ⇒I =∫(−(3/(4t))+((35)/(4(9t−4))))dt =−(3/4)ln∣t∣ +((35)/(36))ln∣9t−4∣ +c −(3/4)(2x)+((35)/(36))ln∣9e^(2x) −4∣ +c =−(3/2)x +((35)/(36))ln∣9e^(2x) −4∣ +c ⇒ A=−(3/2), B =((35)/(36)) and C =c .
$${let}\:{I}\:=\int\:\frac{\mathrm{4}{e}^{{x}} \:+\mathrm{6}{e}^{−{x}} }{\mathrm{9}{e}^{{x}} −\mathrm{4}\:{e}^{−{x}} }{dx}\left[\Rightarrow{I}\:=\int\:\:\frac{\mathrm{4}{e}^{\mathrm{2}{x}} \:+\mathrm{6}}{\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}}{dx}\:\:{changement}\:{e}^{\mathrm{2}{x}} ={t}\:{give}\right. \\ $$$${I}\:=\int\:\:\:\frac{\mathrm{4}{t}+\mathrm{6}}{\mathrm{9}{t}−\mathrm{4}}\:\frac{{dt}}{\mathrm{2}{t}}\:=\int\:\:\:\frac{\mathrm{2}{t}+\mathrm{3}}{{t}\left(\mathrm{9}{t}−\mathrm{4}\right)}{dt}\:{let}\:{decompose}\:{F}\left({t}\right)=\frac{\mathrm{2}{t}+\mathrm{3}}{{t}\left(\mathrm{9}{t}−\mathrm{4}\right)} \\ $$$${F}\left({t}\right)=\frac{{a}}{{t}}\:+\:\frac{{b}}{\mathrm{9}{t}−\mathrm{4}} \\ $$$${a}={lim}_{{t}\rightarrow\mathrm{0}} {tF}\left({t}\right)=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${b}={lim}_{{t}\rightarrow\frac{\mathrm{4}}{\mathrm{9}}} \:\:\:\left(\mathrm{9}{t}−\mathrm{4}\right){F}\left({t}\right)\:=\frac{\mathrm{2}×\frac{\mathrm{4}}{\mathrm{9}}+\mathrm{3}}{\frac{\mathrm{4}}{\mathrm{9}}}\:=\frac{\mathrm{8}+\mathrm{27}}{\mathrm{4}}\:=\frac{\mathrm{35}}{\mathrm{4}}\:\Rightarrow \\ $$$${F}\left({t}\right)=−\frac{\mathrm{3}}{\mathrm{4}{t}}\:+\frac{\mathrm{35}}{\mathrm{4}\left(\mathrm{9}{t}−\mathrm{4}\right)}\:\Rightarrow{I}\:=\int\left(−\frac{\mathrm{3}}{\mathrm{4}{t}}+\frac{\mathrm{35}}{\mathrm{4}\left(\mathrm{9}{t}−\mathrm{4}\right)}\right){dt} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{4}}{ln}\mid{t}\mid\:+\frac{\mathrm{35}}{\mathrm{36}}{ln}\mid\mathrm{9}{t}−\mathrm{4}\mid\:+{c} \\ $$$$−\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{2}{x}\right)+\frac{\mathrm{35}}{\mathrm{36}}{ln}\mid\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\mid\:+{c}\:\:=−\frac{\mathrm{3}}{\mathrm{2}}{x}\:+\frac{\mathrm{35}}{\mathrm{36}}{ln}\mid\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\mid\:+{c}\:\Rightarrow \\ $$$${A}=−\frac{\mathrm{3}}{\mathrm{2}},\:\:\:\:{B}\:=\frac{\mathrm{35}}{\mathrm{36}}\:\:{and}\:\:{C}\:={c}\:. \\ $$

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