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Question Number 53311 by gunawan last updated on 20/Jan/19

If ∫ ((4e^x +6e^(−x) )/(9e^x −4e^(−x) )) dx=Ax+B log(9e^(2x) −4)+C then A=... B=... C=...

$$\mathrm{If}\:\int\:\frac{\mathrm{4}{e}^{{x}} +\mathrm{6}{e}^{−{x}} }{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} }\:{dx}={Ax}+{B}\:\mathrm{log}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)+{C} \\ $$$$\mathrm{then} \\ $$$${A}=... \\ $$$${B}=... \\ $$$${C}=... \\ $$

Commented by maxmathsup by imad last updated on 20/Jan/19

let I =∫ ((4e^x +6e^(−x) )/(9e^x −4e^(−x) ))dx ⇒I =_(e^x =t) ∫ ((4t+6t^(−1) )/(9t−4t^(−1) )) (dt/t) =∫ ((4t+6t^(−1) )/(9t^2 −4))dt =∫ ((4t^2 +6)/(9t^3 −4t)) dt let decompose F(t)=((4t^2 +6)/(t(9t^2 −4))) =((4t^2 +6)/(t(3t−2)(3t+2))) =(a/t) +(b/(3t−2)) +(c/(3t +2)) a =lim_(t→0) tF(t)=−(3/2) b =lim_(t→(2/3)) (3t−2)F(t)=((4.(4/9)+6)/((2/3)(4))) =(((8/9)+3)/(4/3)) =((35)/9) .(3/4) =((35)/(12)) c =lim_(t→−(2/3)) (3t+2)F(t) =((4.(4/9)+6)/((−(2/3))(−4))) =((35)/(12)) ⇒ I =−(3/2)∫ (dt/t) +((35)/(12)) ∫ (dt/(3t−2)) +((35)/(12)) ∫ (dt/(3t +2)) +c =−(3/2)ln∣t∣ +((35)/(36))ln∣3t−2∣ +((35)/(36))ln∣3t+2∣ +c =−(3/2)ln∣t∣ +((35)/(36))ln∣9t^2 −4∣ +c =−(3/2)x +((35)/(36)) ln(9e^(2x) −4) +c ⇒a=−(3/2) and b=((35)/(36)) c →constant of integration.

$${let}\:{I}\:=\int\:\:\frac{\mathrm{4}{e}^{{x}} \:+\mathrm{6}{e}^{−{x}} }{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} }{dx}\:\Rightarrow{I}\:=_{{e}^{{x}} ={t}} \:\:\int\:\:\:\frac{\mathrm{4}{t}+\mathrm{6}{t}^{−\mathrm{1}} }{\mathrm{9}{t}−\mathrm{4}{t}^{−\mathrm{1}} }\:\frac{{dt}}{{t}}\:=\int\:\:\frac{\mathrm{4}{t}+\mathrm{6}{t}^{−\mathrm{1}} }{\mathrm{9}{t}^{\mathrm{2}} −\mathrm{4}}{dt} \\ $$$$=\int\:\:\frac{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{6}}{\mathrm{9}{t}^{\mathrm{3}} −\mathrm{4}{t}}\:{dt}\:\:{let}\:{decompose}\:{F}\left({t}\right)=\frac{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{6}}{{t}\left(\mathrm{9}{t}^{\mathrm{2}} −\mathrm{4}\right)}\:=\frac{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{6}}{{t}\left(\mathrm{3}{t}−\mathrm{2}\right)\left(\mathrm{3}{t}+\mathrm{2}\right)} \\ $$$$=\frac{{a}}{{t}}\:+\frac{{b}}{\mathrm{3}{t}−\mathrm{2}}\:+\frac{{c}}{\mathrm{3}{t}\:+\mathrm{2}} \\ $$$${a}\:={lim}_{{t}\rightarrow\mathrm{0}} {tF}\left({t}\right)=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${b}\:={lim}_{{t}\rightarrow\frac{\mathrm{2}}{\mathrm{3}}} \:\:\left(\mathrm{3}{t}−\mathrm{2}\right){F}\left({t}\right)=\frac{\mathrm{4}.\frac{\mathrm{4}}{\mathrm{9}}+\mathrm{6}}{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{4}\right)}\:=\frac{\frac{\mathrm{8}}{\mathrm{9}}+\mathrm{3}}{\frac{\mathrm{4}}{\mathrm{3}}}\:=\frac{\mathrm{35}}{\mathrm{9}}\:.\frac{\mathrm{3}}{\mathrm{4}}\:=\frac{\mathrm{35}}{\mathrm{12}} \\ $$$${c}\:={lim}_{{t}\rightarrow−\frac{\mathrm{2}}{\mathrm{3}}} \:\:\:\left(\mathrm{3}{t}+\mathrm{2}\right){F}\left({t}\right)\:=\frac{\mathrm{4}.\frac{\mathrm{4}}{\mathrm{9}}+\mathrm{6}}{\left(−\frac{\mathrm{2}}{\mathrm{3}}\right)\left(−\mathrm{4}\right)}\:=\frac{\mathrm{35}}{\mathrm{12}}\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{3}}{\mathrm{2}}\int\:\frac{{dt}}{{t}}\:+\frac{\mathrm{35}}{\mathrm{12}}\:\int\:\frac{{dt}}{\mathrm{3}{t}−\mathrm{2}}\:+\frac{\mathrm{35}}{\mathrm{12}}\:\int\:\:\frac{{dt}}{\mathrm{3}{t}\:+\mathrm{2}}\:+{c} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}{ln}\mid{t}\mid\:+\frac{\mathrm{35}}{\mathrm{36}}{ln}\mid\mathrm{3}{t}−\mathrm{2}\mid\:+\frac{\mathrm{35}}{\mathrm{36}}{ln}\mid\mathrm{3}{t}+\mathrm{2}\mid\:+{c} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}{ln}\mid{t}\mid\:+\frac{\mathrm{35}}{\mathrm{36}}{ln}\mid\mathrm{9}{t}^{\mathrm{2}} −\mathrm{4}\mid\:+{c} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}{x}\:+\frac{\mathrm{35}}{\mathrm{36}}\:{ln}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)\:+{c}\:\Rightarrow{a}=−\frac{\mathrm{3}}{\mathrm{2}}\:{and}\:{b}=\frac{\mathrm{35}}{\mathrm{36}} \\ $$$${c}\:\rightarrow{constant}\:{of}\:{integration}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

4e^x +6e^(−x) =P(9e^x −4e^(−x) )+Q×(d/dx)(9e^x −4e^(−x) ) =P(9e^x −4e^(−x) )+Q(9e^x +4e^(−x) ) =e^x (9P+9Q)+e^(−x) (−4P+4Q) 9P+9Q=4 ×4 −4P+4Q=6 ×9 36P+36Q=16 −36P+36Q=54 72Q=70 [Q=((35)/(36))] 4P=4×((35)/(36))−6 4P=((140−216)/(36)) [P=((−76)/(4×36))=((−19)/(36))] ∫((4e^x +6e^(−x) )/(9e^x −4e^(−x) ))dx =∫((P(9e^x −4e^(−x) )+Q×(d/dx)(9e^x −4e^(−x) ))/(9e^x −4e^(−x) ))dx =P∫dx+Q∫((d(9e^x −4e^(−x) ))/(9e^x −4e^(−x) )) =Px+Qln(9e^x −4e^(−x) )+c_1 =Px+Q[ln(((9e^(2x) −4)/e^x ))]+c_1 =Px+Qln(9e^(2x) −4)−Qln(e^x )+c_1 =Px+Qln(9e^(2x) −4)−Qx+c_1 =(P−Q)x+Qln(9e^(2x) −4)+c_1 =(((−19)/(36))−((35)/(36)))x+((35)/(36))ln(9e^(2x) −4)+c_1 =(((−54)/(36)))x+((35)/(36))ln(9e^(2x) −4)+c_1 A→((−54)/(36)) B→((35)/(36)) C→c_1 pls check steps mistake if any

$$\mathrm{4}{e}^{{x}} +\mathrm{6}{e}^{−{x}} ={P}\left(\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} \right)+{Q}×\frac{{d}}{{dx}}\left(\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} \right) \\ $$$$={P}\left(\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} \right)+{Q}\left(\mathrm{9}{e}^{{x}} +\mathrm{4}{e}^{−{x}} \right) \\ $$$$={e}^{{x}} \left(\mathrm{9}{P}+\mathrm{9}{Q}\right)+{e}^{−{x}} \left(−\mathrm{4}{P}+\mathrm{4}{Q}\right) \\ $$$$\mathrm{9}{P}+\mathrm{9}{Q}=\mathrm{4}\:\:×\mathrm{4} \\ $$$$−\mathrm{4}{P}+\mathrm{4}{Q}=\mathrm{6}\:×\mathrm{9} \\ $$$$\mathrm{36}{P}+\mathrm{36}{Q}=\mathrm{16} \\ $$$$−\mathrm{36}{P}+\mathrm{36}{Q}=\mathrm{54} \\ $$$$\mathrm{72}{Q}=\mathrm{70}\:\left[{Q}=\frac{\mathrm{35}}{\mathrm{36}}\right]\:\:\:\mathrm{4}{P}=\mathrm{4}×\frac{\mathrm{35}}{\mathrm{36}}−\mathrm{6} \\ $$$$\mathrm{4}{P}=\frac{\mathrm{140}−\mathrm{216}}{\mathrm{36}}\:\:\left[{P}=\frac{−\mathrm{76}}{\mathrm{4}×\mathrm{36}}=\frac{−\mathrm{19}}{\mathrm{36}}\right] \\ $$$$\int\frac{\mathrm{4}{e}^{{x}} +\mathrm{6}{e}^{−{x}} }{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} }{dx} \\ $$$$=\int\frac{{P}\left(\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} \right)+{Q}×\frac{{d}}{{dx}}\left(\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} \right)}{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} }{dx} \\ $$$$={P}\int{dx}+{Q}\int\frac{{d}\left(\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} \right)}{\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} } \\ $$$$={Px}+{Qln}\left(\mathrm{9}{e}^{{x}} −\mathrm{4}{e}^{−{x}} \right)+{c}_{\mathrm{1}} \\ $$$$={Px}+{Q}\left[{ln}\left(\frac{\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}}{{e}^{{x}} }\right)\right]+{c}_{\mathrm{1}} \\ $$$$={Px}+{Qln}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)−{Qln}\left({e}^{{x}} \right)+{c}_{\mathrm{1}} \\ $$$$={Px}+{Qln}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)−{Qx}+{c}_{\mathrm{1}} \\ $$$$=\left({P}−{Q}\right){x}+{Qln}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)+{c}_{\mathrm{1}} \\ $$$$=\left(\frac{−\mathrm{19}}{\mathrm{36}}−\frac{\mathrm{35}}{\mathrm{36}}\right){x}+\frac{\mathrm{35}}{\mathrm{36}}{ln}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)+{c}_{\mathrm{1}} \\ $$$$=\left(\frac{−\mathrm{54}}{\mathrm{36}}\right){x}+\frac{\mathrm{35}}{\mathrm{36}}{ln}\left(\mathrm{9}{e}^{\mathrm{2}{x}} −\mathrm{4}\right)+{c}_{\mathrm{1}} \\ $$$${A}\rightarrow\frac{−\mathrm{54}}{\mathrm{36}}\:\:\:{B}\rightarrow\frac{\mathrm{35}}{\mathrm{36}}\:\:{C}\rightarrow{c}_{\mathrm{1}} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}}\:\boldsymbol{{steps}}\:\boldsymbol{{mistake}}\:\boldsymbol{{if}}\:\boldsymbol{{any}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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