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Question Number 109854 by bemath last updated on 26/Aug/20

 ((★be★)/(math))  (1)∫ ((3+2cos x)/((2+3cos x)^3 )) dx   (2)((√(2+(√3))) )^y −((√(2−(√3))) )^y  = 14       y=?

$$\:\frac{\bigstar{be}\bigstar}{{math}} \\ $$$$\left(\mathrm{1}\right)\int\:\frac{\mathrm{3}+\mathrm{2cos}\:{x}}{\left(\mathrm{2}+\mathrm{3cos}\:{x}\right)^{\mathrm{3}} }\:{dx}\: \\ $$$$\left(\mathrm{2}\right)\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{y}} −\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{y}} \:=\:\mathrm{14} \\ $$$$\:\:\:\:\:{y}=? \\ $$

Commented by bemath last updated on 26/Aug/20

thank you both

$${thank}\:{you}\:{both} \\ $$

Answered by john santu last updated on 26/Aug/20

 consider (√(2+(√3))) =(√(((((2+(√3))(2−(√(3))))/(2−(√3))))))                       =(√(1/(2−(√3)))) = (1/( (√(2−(√3)))))  set ((√(2+(√3))) )^y = t   ⇔ t−(1/t) = 14 ⇒t^2 −1=14t      t^2 −14t−1=0 ,(t−7)^2 −50=0      t= 7+5(√2) ; ((√(2+(√3))) )^y =7+5(√2)      (2+(√3) )^y =(7+5(√2))^2        (2+(√3))^y =99+70(√2)        y = ((ln (99+70(√2)))/(ln (2+(√3))))≈ 4.015497788

$$\:{consider}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:=\sqrt{\left(\frac{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\left.\mathrm{3}\right)}\right.}{\mathrm{2}−\sqrt{\mathrm{3}}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}} \\ $$$${set}\:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{y}} =\:{t}\: \\ $$$$\Leftrightarrow\:{t}−\frac{\mathrm{1}}{{t}}\:=\:\mathrm{14}\:\Rightarrow{t}^{\mathrm{2}} −\mathrm{1}=\mathrm{14}{t} \\ $$$$\:\:\:\:{t}^{\mathrm{2}} −\mathrm{14}{t}−\mathrm{1}=\mathrm{0}\:,\left({t}−\mathrm{7}\right)^{\mathrm{2}} −\mathrm{50}=\mathrm{0} \\ $$$$\:\:\:\:{t}=\:\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\:;\:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{y}} =\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{{y}} =\left(\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{y}} =\mathrm{99}+\mathrm{70}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{y}\:=\:\frac{\mathrm{ln}\:\left(\mathrm{99}+\mathrm{70}\sqrt{\mathrm{2}}\right)}{\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}\approx\:\mathrm{4}.\mathrm{015497788} \\ $$

Answered by Her_Majesty last updated on 26/Aug/20

(√(2+(√3)))=(((√6)+(√2))/2)  (√(2−(√3)))=(((√6)−(√2))/2)=(1/( (√(2+(√3)))))  a^y −(1/a^y )=2sinh(ylna)=14 ⇒ y=((sinh^(−1) 7)/(lna))=  =((2sinh^(−1) 7)/(ln(2+(√3))))=((2ln(7+5(√2)))/(ln(2+(√3))))

$$\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}} \\ $$$${a}^{{y}} −\frac{\mathrm{1}}{{a}^{{y}} }=\mathrm{2}{sinh}\left({ylna}\right)=\mathrm{14}\:\Rightarrow\:{y}=\frac{{sinh}^{−\mathrm{1}} \mathrm{7}}{{lna}}= \\ $$$$=\frac{\mathrm{2}{sinh}^{−\mathrm{1}} \mathrm{7}}{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}=\frac{\mathrm{2}{ln}\left(\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\right)}{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \\ $$$$ \\ $$

Answered by Sarah85 last updated on 26/Aug/20

∫((3+2cos x)/((2+3cos x)^3 ))dx  as usual: t=tan (x/2)  −2∫(((t^2 +1)(t^2 +5))/((t^2 −5)^3 ))dt  I prefer Ostrogradski′s Method I learned from  Sir MJS  ⇒  ((t(7t^2 −5))/(5(t^2 −5)^2 ))−(3/5)∫(dt/(t^2 −5))=((t(7t^2 −5))/(5(t^2 −5)^2 ))−((3(√5))/(50))ln (((√5)t−5)/( (√5)t+5)) =...

$$\int\frac{\mathrm{3}+\mathrm{2cos}\:{x}}{\left(\mathrm{2}+\mathrm{3cos}\:{x}\right)^{\mathrm{3}} }{dx} \\ $$$$\mathrm{as}\:\mathrm{usual}:\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}} \\ $$$$−\mathrm{2}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{5}\right)}{\left({t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{3}} }{dt} \\ $$$$\mathrm{I}\:\mathrm{prefer}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\mathrm{I}\:\mathrm{learned}\:\mathrm{from} \\ $$$$\mathrm{Sir}\:\mathrm{MJS} \\ $$$$\Rightarrow \\ $$$$\frac{{t}\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{5}\right)}{\mathrm{5}\left({t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} }−\frac{\mathrm{3}}{\mathrm{5}}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{5}}=\frac{{t}\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{5}\right)}{\mathrm{5}\left({t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} }−\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{50}}\mathrm{ln}\:\frac{\sqrt{\mathrm{5}}{t}−\mathrm{5}}{\:\sqrt{\mathrm{5}}{t}+\mathrm{5}}\:=... \\ $$

Answered by 1549442205PVT last updated on 26/Aug/20

F=∫((3+2cosx)/((2+3cosx)^3 ))dx.Put t=tan(x/2)  ⇒dt=(1/2)(1+t^2 )dx⇒dx=((2dt)/(1+t^2 ))  2+3cosx=2+3((1−t^2 )/(1+t^2 ))=((−t^2 +5)/(1+t^2 ))  3+2cosx=3+((2(1−t^2 ))/(1+t^2 ))=((t^2 +5)/(1+t^2 ))  F=2∫(((t^2 +5)(t^2 +1)dt)/((−t^2 +5)^3 ))=−2∫(((t^2 +5)(t^2 +1)dt)/((t^2 −5)^3 ))  Since  (t^2 +5)(t^2 +1)=t^4 +6t^2 +5  =(t^2 −5)^2 +16(t^2 −5)+60 ,we have  F=−∫[(2/(t^2 −5))+((32)/((t^2 −5)^2 ))+((60)/((t^2 −5)^3 ))]dt  we have     (1/(t^2 −5))=(1/((t+(√5))(t−(√5))))  =(1/(2(√5)))((1/(t−(√5)))−(1/(t+(√5)))). Put a= (1/(t−(√5)))  b=(1/(t+(√5)))⇒ab=(1/(2(√5)))(a−b).Hence,  (1/((t^2 −5)^2 ))=(ab)^2 =(1/(20))(a^2 +b^2 −(1/( (√5)))(a−b))  (1/((t^2 −5)^3 ))=(ab)^3 =(1/(40(√5)))(a^3 −b^3 −3ab(a−b))  =(1/(40(√5)))(a^3 −b^3 −(3/(2(√5)))(a−b)^2 )  =(1/(40(√5)))(a^3 −b^3 )−(3/(400))(a^2 +b^2 −(1/( (√5)))(a−b))  ∫a^3 dt=∫(dt/((t−(√5))^3 ))=((−1)/(2(t−(√5))^2 )),∫b^3 dt=((−1)/(2(t+(√5))^2 ))⇒∫(a^3 −b^3 )dt=(1/2)((1/((t+(√5))^2 ))−(1/((t−(√5))^2 )))=((−2t(√5))/((t^2 −5)^2 )) (1)  ∫a^2 dt=∫(dt/((t−(√5))^2 ))=((−1)/(t−(√5))),∫b^2 dt=((−1)/(t+(√5)))⇒∫(a^2 +b^2 )dt=−((1/(t−(√5)))+(1/(t+(√5))))=((−2t)/(t^2 −5))(2)  ∫(a−b)dt=∫((1/(t−(√5)))−(1/(t+(√5))))dt=ln∣((t−(√5))/(t+(√5)))∣(3)  Consequemtly,  2ab+32(ab)^2 +120(ab)^3 =(1/( (√5)))(a−b)+((16)/(10))(a^2 +b^2 )−((16)/(10(√5)))(a−b)+(3/( (√5)))(a^3 −b^3 )−(9/(10))(a^2 +b^2 )+(9/(10(√5)))(a−b)  =(3/( (√5)))(a^3 −b^3 )+(7/(10))(a^2 +b^2 )+(3/(10(√5)))(a−b)  From (1)(2)(3)we get  F=−{(3/( (√5)))×((2t(√5))/((t^2 −5)^2 ))+(7/(10))×((−2t)/(t^2 −5))+(3/(10(√5)))×ln∣((t−(√5))/(t+(√5)))∣}  =((6t)/((t^2 −5)^2 ))+((7t)/(5(t^2 −5)))−(3/(10(√5)))ln∣((t−(√5))/(t+(√5)))∣=((t(7t^2 −5))/(5(t^2 −5)))−(3/(10(√5)))ln∣((t−(√5))/(t+(√5)))∣  =((tan(x/2)(7tan^2 (x/2)−5))/(5(tan^2 (x/2)−5)^2 ))−((3(√5))/(50))ln∣((tan(x/2)−(√5))/(tan(x/2)+(√5)))∣  2)Since (√(2+(√3))) .(√(2−(√3)))=(√((2+(√3))(2−(√3))))  =(√(4−3))=1⇒((√(2+(√3))))^y .((√(2−(√3))))^y =1,so  Put ((√(2+(√3))))^y =x(x>0)we get  ((√(2−(√3))))^y =(1/x).Hence  ((√(2+(√3))) )^y −((√(2−(√3))) )^y  = 14  ⇔x−(1/x)=14⇔x^2 −14x−1=0  Δ′=7^2 +1=50=(5(√2))^2 ⇒x=7+5(√2)  (root x=7−5(√2)<0⇒rejected).Hence  ((√(2+(√3))))^y =7+5(√2) ⇒yln((√(2+(√3))))=ln(7+5(√2))  ⇔y=((ln(7+5(√2)))/(ln((√(2+(√3))))))≈4.0155

$$\mathrm{F}=\int\frac{\mathrm{3}+\mathrm{2cosx}}{\left(\mathrm{2}+\mathrm{3cosx}\right)^{\mathrm{3}} }\mathrm{dx}.\mathrm{Put}\:\mathrm{t}=\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dx}\Rightarrow\mathrm{dx}=\frac{\mathrm{2dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{2}+\mathrm{3cosx}=\mathrm{2}+\mathrm{3}\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }=\frac{−\mathrm{t}^{\mathrm{2}} +\mathrm{5}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{3}+\mathrm{2cosx}=\mathrm{3}+\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }=\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{5}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$\mathrm{F}=\mathrm{2}\int\frac{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{5}\right)\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{dt}}{\left(−\mathrm{t}^{\mathrm{2}} +\mathrm{5}\right)^{\mathrm{3}} }=−\mathrm{2}\int\frac{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{5}\right)\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{3}} } \\ $$$$\mathrm{Since}\:\:\left(\mathrm{t}^{\mathrm{2}} +\mathrm{5}\right)\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{t}^{\mathrm{4}} +\mathrm{6t}^{\mathrm{2}} +\mathrm{5} \\ $$$$=\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} +\mathrm{16}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)+\mathrm{60}\:,\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{F}=−\int\left[\frac{\mathrm{2}}{\mathrm{t}^{\mathrm{2}} −\mathrm{5}}+\frac{\mathrm{32}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} }+\frac{\mathrm{60}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{3}} }\right]\mathrm{dt} \\ $$$$\mathrm{we}\:\mathrm{have}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} −\mathrm{5}}=\frac{\mathrm{1}}{\left(\mathrm{t}+\sqrt{\mathrm{5}}\right)\left(\mathrm{t}−\sqrt{\mathrm{5}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\left(\frac{\mathrm{1}}{\mathrm{t}−\sqrt{\mathrm{5}}}−\frac{\mathrm{1}}{\mathrm{t}+\sqrt{\mathrm{5}}}\right).\:\mathrm{Put}\:\mathrm{a}=\:\frac{\mathrm{1}}{\mathrm{t}−\sqrt{\mathrm{5}}} \\ $$$$\mathrm{b}=\frac{\mathrm{1}}{\mathrm{t}+\sqrt{\mathrm{5}}}\Rightarrow\mathrm{ab}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\left(\mathrm{a}−\mathrm{b}\right).\mathrm{Hence}, \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} }=\left(\mathrm{ab}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{20}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{a}−\mathrm{b}\right)\right) \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{3}} }=\left(\mathrm{ab}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{40}\sqrt{\mathrm{5}}}\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} −\mathrm{3ab}\left(\mathrm{a}−\mathrm{b}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{40}\sqrt{\mathrm{5}}}\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{5}}}\left(\mathrm{a}−\mathrm{b}\right)^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{40}\sqrt{\mathrm{5}}}\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} \right)−\frac{\mathrm{3}}{\mathrm{400}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{a}−\mathrm{b}\right)\right) \\ $$$$\int\mathrm{a}^{\mathrm{3}} \mathrm{dt}=\int\frac{\mathrm{dt}}{\left(\mathrm{t}−\sqrt{\mathrm{5}}\right)^{\mathrm{3}} }=\frac{−\mathrm{1}}{\mathrm{2}\left(\mathrm{t}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} },\int\mathrm{b}^{\mathrm{3}} \mathrm{dt}=\frac{−\mathrm{1}}{\mathrm{2}\left(\mathrm{t}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }\Rightarrow\int\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} \right)\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\left(\mathrm{t}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\mathrm{t}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }\right)=\frac{−\mathrm{2t}\sqrt{\mathrm{5}}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} }\:\left(\mathrm{1}\right) \\ $$$$\int\mathrm{a}^{\mathrm{2}} \mathrm{dt}=\int\frac{\mathrm{dt}}{\left(\mathrm{t}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }=\frac{−\mathrm{1}}{\mathrm{t}−\sqrt{\mathrm{5}}},\int\mathrm{b}^{\mathrm{2}} \mathrm{dt}=\frac{−\mathrm{1}}{\mathrm{t}+\sqrt{\mathrm{5}}}\Rightarrow\int\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\mathrm{dt}=−\left(\frac{\mathrm{1}}{\mathrm{t}−\sqrt{\mathrm{5}}}+\frac{\mathrm{1}}{\mathrm{t}+\sqrt{\mathrm{5}}}\right)=\frac{−\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} −\mathrm{5}}\left(\mathrm{2}\right) \\ $$$$\int\left(\mathrm{a}−\mathrm{b}\right)\mathrm{dt}=\int\left(\frac{\mathrm{1}}{\mathrm{t}−\sqrt{\mathrm{5}}}−\frac{\mathrm{1}}{\mathrm{t}+\sqrt{\mathrm{5}}}\right)\mathrm{dt}=\mathrm{ln}\mid\frac{\mathrm{t}−\sqrt{\mathrm{5}}}{\mathrm{t}+\sqrt{\mathrm{5}}}\mid\left(\mathrm{3}\right) \\ $$$$\mathrm{Consequemtly}, \\ $$$$\mathrm{2ab}+\mathrm{32}\left(\mathrm{ab}\right)^{\mathrm{2}} +\mathrm{120}\left(\mathrm{ab}\right)^{\mathrm{3}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{a}−\mathrm{b}\right)+\frac{\mathrm{16}}{\mathrm{10}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)−\frac{\mathrm{16}}{\mathrm{10}\sqrt{\mathrm{5}}}\left(\mathrm{a}−\mathrm{b}\right)+\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} \right)−\frac{\mathrm{9}}{\mathrm{10}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)+\frac{\mathrm{9}}{\mathrm{10}\sqrt{\mathrm{5}}}\left(\mathrm{a}−\mathrm{b}\right) \\ $$$$=\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}}}\left(\mathrm{a}^{\mathrm{3}} −\mathrm{b}^{\mathrm{3}} \right)+\frac{\mathrm{7}}{\mathrm{10}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)+\frac{\mathrm{3}}{\mathrm{10}\sqrt{\mathrm{5}}}\left(\mathrm{a}−\mathrm{b}\right) \\ $$$$\mathrm{From}\:\left(\mathrm{1}\right)\left(\mathrm{2}\right)\left(\mathrm{3}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{F}=−\left\{\frac{\mathrm{3}}{\:\sqrt{\mathrm{5}}}×\frac{\mathrm{2t}\sqrt{\mathrm{5}}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} }+\frac{\mathrm{7}}{\mathrm{10}}×\frac{−\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} −\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{10}\sqrt{\mathrm{5}}}×\mathrm{ln}\mid\frac{\mathrm{t}−\sqrt{\mathrm{5}}}{\mathrm{t}+\sqrt{\mathrm{5}}}\mid\right\} \\ $$$$=\frac{\mathrm{6t}}{\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)^{\mathrm{2}} }+\frac{\mathrm{7t}}{\mathrm{5}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)}−\frac{\mathrm{3}}{\mathrm{10}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\frac{\mathrm{t}−\sqrt{\mathrm{5}}}{\mathrm{t}+\sqrt{\mathrm{5}}}\mid=\frac{\mathrm{t}\left(\mathrm{7t}^{\mathrm{2}} −\mathrm{5}\right)}{\mathrm{5}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{5}\right)}−\frac{\mathrm{3}}{\mathrm{10}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\frac{\mathrm{t}−\sqrt{\mathrm{5}}}{\mathrm{t}+\sqrt{\mathrm{5}}}\mid \\ $$$$=\frac{\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\left(\mathrm{7}\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\mathrm{5}\right)}{\mathrm{5}\left(\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\mathrm{5}\right)^{\mathrm{2}} }−\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{50}}\boldsymbol{\mathrm{ln}}\mid\frac{\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}−\sqrt{\mathrm{5}}}{\boldsymbol{\mathrm{tan}}\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}+\sqrt{\mathrm{5}}}\mid \\ $$$$\left.\mathrm{2}\left.\right)\mathrm{Since}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:.\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}=\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}}\right.}\right) \\ $$$$=\sqrt{\mathrm{4}−\mathrm{3}}=\mathrm{1}\Rightarrow\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{\mathrm{y}} .\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{\mathrm{y}} =\mathrm{1},\mathrm{so} \\ $$$$\mathrm{Put}\:\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{\mathrm{y}} =\mathrm{x}\left(\mathrm{x}>\mathrm{0}\right)\mathrm{we}\:\mathrm{get} \\ $$$$\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right)^{\mathrm{y}} =\frac{\mathrm{1}}{\mathrm{x}}.\mathrm{Hence} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{y}} −\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{y}} \:=\:\mathrm{14} \\ $$$$\Leftrightarrow\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{14}\Leftrightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{14x}−\mathrm{1}=\mathrm{0} \\ $$$$\Delta'=\mathrm{7}^{\mathrm{2}} +\mathrm{1}=\mathrm{50}=\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \Rightarrow\mathrm{x}=\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{root}\:\mathrm{x}=\mathrm{7}−\mathrm{5}\sqrt{\mathrm{2}}<\mathrm{0}\Rightarrow\mathrm{rejected}\right).\mathrm{Hence} \\ $$$$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)^{\mathrm{y}} =\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{yln}\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)=\mathrm{ln}\left(\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\right) \\ $$$$\Leftrightarrow\boldsymbol{\mathrm{y}}=\frac{\boldsymbol{\mathrm{ln}}\left(\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}\right)}{\boldsymbol{\mathrm{ln}}\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\right)}\approx\mathrm{4}.\mathrm{0155} \\ $$

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