Question Number 44173 by abdo.msup.com last updated on 22/Sep/18
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\mathrm{3}+{t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{t}}}{dt} \\ $$
Commented by maxmathsup by imad last updated on 23/Sep/18
![let I =∫_0 ^∞ (dt/((3+t^2 )(√(1+t)))) changement (√(1+t))=x give 1+t =x^2 and I =∫_0 ^∞ ((2xdx)/((3+(x^2 −1)^2 )x)) =∫_0 ^∞ ((2dx)/(3 +x^4 −2x^2 +1)) =∫_0 ^∞ ((2dx)/(x^4 −2x^2 +4)) =∫_0 ^∞ ((2dx)/((x^4 +4 −2x^2 ))) =∫_0 ^∞ ((2dx)/((x^2 +2)^2 −6x^2 )) =∫_0 ^∞ ((2dx)/((x^2 +2 +(√6)x)(x^(2 ) +2−(√6)x))) =∫_0 ^∞ ((2dx)/((x^2 +x(√6)+2)(x^2 −(√6)x +2))) let decompose F(x)= (2/((x^2 +x(√6)+2)(x^2 −(√6)x +2))) ⇒F(x)=((ax+b)/(x^2 +x(√6)+2)) +((cx+d)/(x^2 −x(√6)+2)) F(−x) =F(x) ⇒((−ax +b)/(x^2 −x(√6)+2)) +((−cx +d)/(x^2 +x(√6)+2)) =F(x)⇒c=−a and b=d ⇒F(x)= ((ax+b)/(x^2 +x(√6)+2)) +((−ax +b)/(x^2 −x(√6)+2)) F(0) =(1/2) =(b/2) +(b/2) =b ⇒b=(1/2) F(1) =((a+(1/2))/(3+(√6))) +((−a +(1/2))/(3−(√6))) =(2/3) ⇒((2a+1)/(2(3+(√6)))) +(1/2)((−2a+1)/(3−(√6))) =(2/3) ⇒ ((2a+1)/(3+(√6))) +((−2a+1)/(3−(√6))) =(4/3) ⇒((2a(3−(√6)) +3−(√6)−2a(3+(√6)) +3+(√6))/3) =(4/3) ⇒ −4a(√6)+6 =4 ⇒−4a(√6)=−2 ⇒a =(1/(2(√6))) ⇒ F(x) =(((1/(2(√6)))x+(1/2))/(x^2 +x(√6)+2)) +((−(1/(2(√6)))x+(1/2))/(x^2 −x(√6)+2)) =(1/(2(√6))){ ((x+(√6))/(x^2 +x(√6)+2)) +((−x+(√6))/(x^2 −x(√6)+2))} ⇒ 2(√6) ∫ F(x)= (1/2)∫ ((2x+2(√6))/(x^2 +x(√6)+2))dx −(1/2) ∫ ((2x−2(√6))/(x^2 −x(√6)+2))dx =(1/2)ln∣((x^2 +x(√6)+2)/(x^2 −x(√6)+2))∣ +(√6)∫ (dx/(x^2 +x(√6)+2)) +(√6) ∫ (dx/(x^2 −x(√6)+2)) but ∫ (dx/(x^2 +x(√6)+2)) =∫ (dx/(x^2 +2((√6)/2)x +(3/2) +2−(3/2))) =∫ (dx/((x+((√6)/2))^2 +(1/2))) =_(x+((√6)/2)=(u/( (√2)))) 2∫ (1/(1+u^2 )) (du/( (√2))) =(√2)arctan (x(√2)+(√3))+c_1 also ∫ (dx/(x^2 −x(√6)+2)) =(√2)arctan(x(√2)−(√3)) +c ⇒ 2(√6)∫_0 ^(+∞) F(x)dx =[(1/2)ln∣((x^2 +x(√6)+2)/(x^2 −x(√6)+2))∣]_0 ^(+∞) +(√6)[(√2)arctan(x(√2)+(√3))]_0 ^(+∞) +(√6)[(√2)arctan(x(√2)−(√3))]_0 ^(+∞) =(√6){(π/2)(√2)−(√2)arctan((√3))}+(√6){(π/2)(√2)+(√2)arctan((√3))} =π(√3) +π(√3) =2π(√3) ⇒∫_0 ^∞ F(x)dx =((2π(√3))/(2(√6))) =(π/( (√2))) ⇒ I = (π/( (√2))) .](https://www.tinkutara.com/question/Q44195.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left(\mathrm{3}+{t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{t}}}\:\:{changement}\:\sqrt{\mathrm{1}+{t}}={x}\:{give}\:\mathrm{1}+{t}\:={x}^{\mathrm{2}} \:{and} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{xdx}}{\left(\mathrm{3}+\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \right){x}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dx}}{\mathrm{3}\:+{x}^{\mathrm{4}} \:−\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dx}}{{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dx}}{\left({x}^{\mathrm{4}} \:+\mathrm{4}\:−\mathrm{2}{x}^{\mathrm{2}} \right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} −\mathrm{6}{x}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}\:+\sqrt{\mathrm{6}}{x}\right)\left({x}^{\mathrm{2}\:} +\mathrm{2}−\sqrt{\mathrm{6}}{x}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{dx}}{\left({x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{6}}{x}\:+\mathrm{2}\right)}\:\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\sqrt{\mathrm{6}}{x}\:+\mathrm{2}\right)}\:\Rightarrow{F}\left({x}\right)=\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:−{x}\sqrt{\mathrm{6}}+\mathrm{2}} \\ $$$${F}\left(−{x}\right)\:={F}\left({x}\right)\:\Rightarrow\frac{−{ax}\:+{b}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:+\frac{−{cx}\:+{d}}{{x}^{\mathrm{2}} +{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:={F}\left({x}\right)\Rightarrow{c}=−{a}\:{and}\:{b}={d} \\ $$$$\Rightarrow{F}\left({x}\right)=\:\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:+\frac{−{ax}\:+{b}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{6}}+\mathrm{2}} \\ $$$${F}\left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{{b}}{\mathrm{2}}\:+\frac{{b}}{\mathrm{2}}\:={b}\:\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{{a}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{6}}}\:+\frac{−{a}\:+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{3}−\sqrt{\mathrm{6}}}\:=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{2}{a}+\mathrm{1}}{\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{6}}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}}\frac{−\mathrm{2}{a}+\mathrm{1}}{\mathrm{3}−\sqrt{\mathrm{6}}}\:=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow \\ $$$$\frac{\mathrm{2}{a}+\mathrm{1}}{\mathrm{3}+\sqrt{\mathrm{6}}}\:+\frac{−\mathrm{2}{a}+\mathrm{1}}{\mathrm{3}−\sqrt{\mathrm{6}}}\:=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{2}{a}\left(\mathrm{3}−\sqrt{\mathrm{6}}\right)\:+\mathrm{3}−\sqrt{\mathrm{6}}−\mathrm{2}{a}\left(\mathrm{3}+\sqrt{\mathrm{6}}\right)\:+\mathrm{3}+\sqrt{\mathrm{6}}}{\mathrm{3}}\:=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow \\ $$$$−\mathrm{4}{a}\sqrt{\mathrm{6}}+\mathrm{6}\:=\mathrm{4}\:\Rightarrow−\mathrm{4}{a}\sqrt{\mathrm{6}}=−\mathrm{2}\:\Rightarrow{a}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}}}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=\frac{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}}}{x}+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:\:+\frac{−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}}}{x}+\frac{\mathrm{1}}{\mathrm{2}}}{{x}^{\mathrm{2}} \:−{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{6}}}\left\{\:\:\frac{{x}+\sqrt{\mathrm{6}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:+\frac{−{x}+\sqrt{\mathrm{6}}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{6}}+\mathrm{2}}\right\}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{\mathrm{6}}\:\int\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\frac{\mathrm{2}{x}+\mathrm{2}\sqrt{\mathrm{6}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}}{dx}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\mathrm{2}{x}−\mathrm{2}\sqrt{\mathrm{6}}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{6}}+\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{6}}+\mathrm{2}}\mid\:+\sqrt{\mathrm{6}}\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:+\sqrt{\mathrm{6}}\:\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:{but} \\ $$$$\int\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} +{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:=\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}{x}\:+\frac{\mathrm{3}}{\mathrm{2}}\:+\mathrm{2}−\frac{\mathrm{3}}{\mathrm{2}}}\:=\int\:\:\:\frac{{dx}}{\left({x}+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=_{{x}+\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}=\frac{{u}}{\:\sqrt{\mathrm{2}}}} \:\:\:\:\:\mathrm{2}\int\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{{du}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}}{arctan}\:\left({x}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)+{c}_{\mathrm{1}} \:\:{also} \\ $$$$\int\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}\sqrt{\mathrm{6}}+\mathrm{2}}\:=\sqrt{\mathrm{2}}{arctan}\left({x}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)\:+{c}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{\mathrm{6}}\int_{\mathrm{0}} ^{+\infty} {F}\left({x}\right){dx}\:=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{{x}^{\mathrm{2}} \:+{x}\sqrt{\mathrm{6}}+\mathrm{2}}{{x}^{\mathrm{2}} \:−{x}\sqrt{\mathrm{6}}+\mathrm{2}}\mid\right]_{\mathrm{0}} ^{+\infty} \:+\sqrt{\mathrm{6}}\left[\sqrt{\mathrm{2}}{arctan}\left({x}\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}\right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$+\sqrt{\mathrm{6}}\left[\sqrt{\mathrm{2}}{arctan}\left({x}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}\right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\sqrt{\mathrm{6}}\left\{\frac{\pi}{\mathrm{2}}\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{3}}\right)\right\}+\sqrt{\mathrm{6}}\left\{\frac{\pi}{\mathrm{2}}\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{3}}\right)\right\} \\ $$$$=\pi\sqrt{\mathrm{3}}\:\:+\pi\sqrt{\mathrm{3}}\:=\mathrm{2}\pi\sqrt{\mathrm{3}}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{F}\left({x}\right){dx}\:=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{6}}}\:=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\Rightarrow\:{I}\:=\:\frac{\pi}{\:\sqrt{\mathrm{2}}}\:. \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Sep/18
![y^2 =1+t dt=2ydy ∫_1 ^∞ ((2ydy)/(y(3+y^4 −2y^2 +1))) ∫_1 ^∞ ((2dy)/((y^4 −2y^2 +4))) ∫_1 ^∞ ((2/y^2 )/((y^2 −2+(4/y^2 ))))dy d(y+(2/y))=1−(2/y^2 ) d(y−(2/y))=1+(2/y^2 ) (1/2)∫_1 ^∞ (((1+(2/y^2 ))−(1−(2/y^2 )))/((y^2 +(4/y^2 )−2)))dy (1/2)∫_1 ^∞ ((d(y−(2/y))−d(y+(2/y)))/((y^2 +(4/y^2 )−2))) (1/2)[∫_1 ^∞ ((d(y−(2/y)))/((y−(2/y))^2 +2.y.(2/y)−2))−∫_1 ^∞ ((d(y+(2/y)))/((y+(2/y))^2 −2.y.(2/y)+2))] (1/2)[∫_(−1) ^∞ (dx_1 /(x_1 ^2 +2))+∫_3 ^∞ (dx_2 /(x_2 ^2 −2))] now use formula (1/2)[(1/( (√2) ))∣tan^(−1) ((x_1 /( (√2) )))∣_(−1) ^∞ +(1/(2(√2) ))∣ln(((x_2 −(√2) )/(x_2 +(√2))))∣_3 ^∞ ] (1/2)[(1/( (√2)))((Π/2)−{tan^(−1) (((−1)/( (√2) )))}+(1/(2(√2)))∣ln(((1−((√2)/x_2 ))/(1+((√2)/x_2 ))))∣_3 ^∞ ] (1/2)[(1/( (√2))){((Π/2)+tan^(−1) ((1/( (√2))))}+(1/(2(√2)))(0−ln∣(((3−(√2))/(3+(√2)))∣)] pls check...](https://www.tinkutara.com/question/Q44186.png)
$${y}^{\mathrm{2}} =\mathrm{1}+{t}\:\:\:{dt}=\mathrm{2}{ydy} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2}{ydy}}{{y}\left(\mathrm{3}+{y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{2}{dy}}{\left({y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{\frac{\mathrm{2}}{{y}^{\mathrm{2}} }}{\left({y}^{\mathrm{2}} −\mathrm{2}+\frac{\mathrm{4}}{{y}^{\mathrm{2}} }\right)}{dy} \\ $$$${d}\left({y}+\frac{\mathrm{2}}{{y}}\right)=\mathrm{1}−\frac{\mathrm{2}}{{y}^{\mathrm{2}} } \\ $$$${d}\left({y}−\frac{\mathrm{2}}{{y}}\right)=\mathrm{1}+\frac{\mathrm{2}}{{y}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \:\frac{\left(\mathrm{1}+\frac{\mathrm{2}}{{y}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\mathrm{2}}{{y}^{\mathrm{2}} }\right)}{\left({y}^{\mathrm{2}} +\frac{\mathrm{4}}{{y}^{\mathrm{2}} }−\mathrm{2}\right)}{dy} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \frac{{d}\left({y}−\frac{\mathrm{2}}{{y}}\right)−{d}\left({y}+\frac{\mathrm{2}}{{y}}\right)}{\left({y}^{\mathrm{2}} +\frac{\mathrm{4}}{{y}^{\mathrm{2}} }−\mathrm{2}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\int_{\mathrm{1}} ^{\infty} \frac{{d}\left({y}−\frac{\mathrm{2}}{{y}}\right)}{\left({y}−\frac{\mathrm{2}}{{y}}\right)^{\mathrm{2}} +\mathrm{2}.{y}.\frac{\mathrm{2}}{{y}}−\mathrm{2}}−\int_{\mathrm{1}} ^{\infty} \frac{{d}\left({y}+\frac{\mathrm{2}}{{y}}\right)}{\left({y}+\frac{\mathrm{2}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}.{y}.\frac{\mathrm{2}}{{y}}+\mathrm{2}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\int_{−\mathrm{1}} ^{\infty} \:\frac{{dx}_{\mathrm{1}} }{{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}}+\int_{\mathrm{3}} ^{\infty} \frac{{dx}_{\mathrm{2}} }{{x}_{\mathrm{2}} ^{\mathrm{2}} −\mathrm{2}}\right]\:{now}\:{use}\:{formula} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\mid{tan}^{−\mathrm{1}} \left(\frac{{x}_{\mathrm{1}} }{\:\sqrt{\mathrm{2}}\:}\right)\mid_{−\mathrm{1}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}\mid{ln}\left(\frac{{x}_{\mathrm{2}} −\sqrt{\mathrm{2}}\:}{{x}_{\mathrm{2}} +\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{3}} ^{\infty} \:\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\Pi}{\mathrm{2}}−\left\{{tan}^{−\mathrm{1}} \left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\right)\right\}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mid{ln}\left(\frac{\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{{x}_{\mathrm{2}} }}{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{{x}_{\mathrm{2}} }}\right)\mid_{\mathrm{3}} ^{\infty} \right]\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left\{\left(\frac{\Pi}{\mathrm{2}}+{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right\}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{0}−{ln}\mid\left(\frac{\mathrm{3}−\sqrt{\mathrm{2}}}{\mathrm{3}+\sqrt{\mathrm{2}}}\mid\right)\right]\right.\right. \\ $$$${pls}\:{check}… \\ $$$$ \\ $$