Question Number 61978 by maxmathsup by imad last updated on 13/Jun/19
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{xt}^{\mathrm{3}} \right){dt}\:\:{with}\:\:\mid{x}\mid<\mathrm{1} \\ $$ $$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$ $$\left.\mathrm{2}\right){calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{t}^{\mathrm{3}} \right){dt} \\ $$ $$\left.\mathrm{3}\right)\:{calculate}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{sin}\theta\:{t}^{\mathrm{3}} \right){dt}\:\:{with}\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$
Commented bymaxmathsup by imad last updated on 14/Jun/19
![1) we have f^′ (x) =−∫_0 ^1 (t^3 /(1−xt^3 ))dt =(1/x) ∫_0 ^1 ((1−xt^3 −1)/(1−xt^3 )) dt =(1/x) −(1/x)∫_0 ^1 (dt/(1−xt^3 )) let α =^3 (√x) ⇒ ∫_0 ^1 (dt/(1−xt^3 )) =∫_0 ^1 (dt/(1−(αt)^3 )) =_(αt =u) ∫_0 ^α (du/(α(1−u^3 ))) =−(1/α) ∫_0 ^α (du/(u^3 −1)) let decompose F(u) =(1/(u^3 −1)) =(1/((u−1)(u^2 +u+1))) =(a/(u−1)) +((bu +c)/(u^2 +u+1)) a =lim_(u→1) (u−1)F(u) =(1/3) lim_(u→+∞) uF(u) =0 =a+b ⇒b =−(1/3) ⇒F(u)=(1/(3(u−1))) +((−(1/3)u +c)/(u^2 +u+1)) F(0) =−1 =−(1/3) +c ⇒c =−1+(1/3) =−(2/3) ⇒F(u)=(1/(3(u−1))) −(1/3) ((u +2)/(u^2 +u +1)) ⇒ ∫_0 ^α (du/(u^3 −1)) =∫_0 ^α (du/(3(u−1))) −(1/6) ∫_0 ^α ((2u +1+3)/(u^2 +u+1)) = [(1/6)ln((((u−1)^2 )/(u^2 +u+1)))]_0 ^α −(1/2) ∫_0 ^α (du/(u^2 +u +1)) ∫_0 ^α (du/(u^2 +u +1)) =∫_0 ^α (du/((u+(1/2))^2 +(3/4))) =_(u+(1/2)=((√3)/2) z) (4/3) ∫_(1/(√3)) ^((2α+1)/(√3)) (1/(1+z^2 )) ((√3)/2)dz =(2/(√3)){ arctan(((2α +1)/(√3)))−arctan((1/(√3))) ⇒ ∫_0 ^α (du/(u^3 −1)) =(1/6)ln(((α^2 −2α +1)/(α^2 +α +1)))−(1/(√3)) arctan(((2α+1)/(√3)))+(1/2) arctan((1/(√3))) ⇒ f^′ (x) =(1/x) +(1/(xα)){(1/6)ln(((α^2 −2α +1)/(α^2 +α +1)))−(1/(√3)) arctan(((2α+1)/(√3)))+(1/2)arctan((1/(√3)))} =A(x) and α =^3 (√x) ⇒ f(x) =∫ A(x)dx +C ....be continued...](Q62026.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{\mathrm{3}} }{\mathrm{1}−{xt}^{\mathrm{3}} }{dt}\:=\frac{\mathrm{1}}{{x}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−{xt}^{\mathrm{3}} −\mathrm{1}}{\mathrm{1}−{xt}^{\mathrm{3}} }\:{dt} \\ $$ $$=\frac{\mathrm{1}}{{x}}\:−\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}−{xt}^{\mathrm{3}} }\:\:{let}\:\alpha\:=^{\mathrm{3}} \sqrt{{x}}\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}−{xt}^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{1}−\left(\alpha{t}\right)^{\mathrm{3}} } \\ $$ $$=_{\alpha{t}\:={u}} \:\:\:\:\:\int_{\mathrm{0}} ^{\alpha} \:\:\:\frac{{du}}{\alpha\left(\mathrm{1}−{u}^{\mathrm{3}} \right)}\:=−\frac{\mathrm{1}}{\alpha}\:\int_{\mathrm{0}} ^{\alpha} \:\:\frac{{du}}{{u}^{\mathrm{3}} −\mathrm{1}}\:{let}\:{decompose} \\ $$ $${F}\left({u}\right)\:=\frac{\mathrm{1}}{{u}^{\mathrm{3}} −\mathrm{1}}\:=\frac{\mathrm{1}}{\left({u}−\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+{u}+\mathrm{1}\right)}\:=\frac{{a}}{{u}−\mathrm{1}}\:+\frac{{bu}\:+{c}}{{u}^{\mathrm{2}} \:+{u}+\mathrm{1}} \\ $$ $${a}\:={lim}_{{u}\rightarrow\mathrm{1}} \left({u}−\mathrm{1}\right){F}\left({u}\right)\:=\frac{\mathrm{1}}{\mathrm{3}} \\ $$ $${lim}_{{u}\rightarrow+\infty} {uF}\left({u}\right)\:=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}\:=−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow{F}\left({u}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({u}−\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{3}}{u}\:+{c}}{{u}^{\mathrm{2}} \:+{u}+\mathrm{1}} \\ $$ $${F}\left(\mathrm{0}\right)\:=−\mathrm{1}\:=−\frac{\mathrm{1}}{\mathrm{3}}\:+{c}\:\Rightarrow{c}\:=−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\:=−\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow{F}\left({u}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({u}−\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{3}}\:\frac{{u}\:+\mathrm{2}}{{u}^{\mathrm{2}} \:+{u}\:+\mathrm{1}}\:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\alpha} \:\frac{{du}}{{u}^{\mathrm{3}} −\mathrm{1}}\:=\int_{\mathrm{0}} ^{\alpha} \:\:\frac{{du}}{\mathrm{3}\left({u}−\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{6}}\:\int_{\mathrm{0}} ^{\alpha} \:\frac{\mathrm{2}{u}\:+\mathrm{1}+\mathrm{3}}{{u}^{\mathrm{2}} \:+{u}+\mathrm{1}} \\ $$ $$=\:\left[\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\frac{\left({u}−\mathrm{1}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} \:+{u}+\mathrm{1}}\right)\right]_{\mathrm{0}} ^{\alpha} −\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\alpha} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+{u}\:+\mathrm{1}} \\ $$ $$\int_{\mathrm{0}} ^{\alpha} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+{u}\:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\alpha} \:\frac{{du}}{\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{u}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{z}} \:\:\:\:\:\frac{\mathrm{4}}{\mathrm{3}}\:\int_{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}\alpha+\mathrm{1}}{\sqrt{\mathrm{3}}}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{z}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dz} \\ $$ $$=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\left\{\:{arctan}\left(\frac{\mathrm{2}\alpha\:+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:\Rightarrow\right. \\ $$ $$\int_{\mathrm{0}} ^{\alpha} \:\:\frac{{du}}{{u}^{\mathrm{3}} −\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\frac{\alpha^{\mathrm{2}} −\mathrm{2}\alpha\:+\mathrm{1}}{\alpha^{\mathrm{2}} +\alpha\:+\mathrm{1}}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}\alpha+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:\Rightarrow \\ $$ $${f}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{{x}}\:+\frac{\mathrm{1}}{{x}\alpha}\left\{\frac{\mathrm{1}}{\mathrm{6}}{ln}\left(\frac{\alpha^{\mathrm{2}} −\mathrm{2}\alpha\:+\mathrm{1}}{\alpha^{\mathrm{2}} \:+\alpha\:+\mathrm{1}}\right)−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}\alpha+\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right\}\:={A}\left({x}\right) \\ $$ $${and}\:\alpha\:=^{\mathrm{3}} \sqrt{{x}}\:\Rightarrow\:{f}\left({x}\right)\:=\int\:{A}\left({x}\right){dx}\:+{C}\:\:\:\:....{be}\:{continued}... \\ $$
Commented bymaxmathsup by imad last updated on 14/Jun/19
![let use another way let α =^3 (√x) ⇒ f(x) =∫_0 ^1 ln(1−(αt)^3 )dt =∫_0 ^1 ln{(1−αt)(1+αt +α^2 t^2 )}dt =∫_0 ^1 ln(1−αt)dt +∫_0 ^1 ln(α^2 t^2 +αt +1)dt changement 1−αt =u give ∫_0 ^1 ln(1−αt)dt =((−1)/α) ∫_1 ^(1−α) ln(u) du =−(1/α)[ulnu−u]_1 ^(1−α) =−(1/α){ (1−α)ln(1−α)−(1−α) +1} =−(1/α){ (1−α)ln(1−α) +α} =((α−1)/α)ln(1−α)−1 . by parts ∫_0 ^1 ln(α^2 t^2 +αt +1)dt =[t ln(α^2 t^2 +αt +1)]_0 ^1 −∫_0 ^1 t ((2α^2 t +α)/(α^2 t^2 +αt +1)) dt =ln(α^2 +α +1) −∫_0 ^1 ((2α^2 t^2 +αt)/(α^2 t^2 +αt +1)) dt ∫_0 ^1 ((2α^2 t^2 +αt)/(α^2 t^2 +αt +1)) dt =∫_0 ^1 ((2(α^2 t^2 +αt +1)−2αt−2 +αt)/(α^2 t^2 +αt +1)) dt =2 −∫_0 ^1 ((αt +2)/(α^2 t^2 +αt +1)) dt =2−(1/(2α)) ∫_0 ^1 ((2α^2 t +α +3α)/(α^2 t^2 +αt +1)) dt =2−(1/(2α))[ln(α^2 t^2 +αt+1)]_0 ^1 −(3/2) ∫_0 ^1 (dt/(α^2 t^2 +αt +1)) ∫_0 ^1 (dt/(α^2 t^2 +αt +1)) =(1/α^2 ) ∫_0 ^1 (dt/(t^2 +(t/α) +(1/α^2 ))) =(1/α^2 ) ∫_0 ^1 (dt/(t^2 ++((2t)/(2α)) +(1/(4α^2 )) +(1/α^2 ) −(1/(4α^2 )))) =(1/α^2 ) ∫_0 ^1 (dt/((t +(1/(2α)))^2 +(3/α^2 ))) =_(t+(1/(2α))=((√3)/α) x) (1/α^2 ) ∫_(1/(2(√3))) ^((2α+1)/(2(√3))) (α^2 /3) (1/(1+x^2 )) ((√3)/α) dx (we take x>0) =(1/(α(√3))) { arctan(((2α+1)/(2(√3))))−arctan((1/(2(√3))))} ⇒ ∫_0 ^1 ((2α^2 t^2 +αt)/(α^2 t^2 +αt+1)) dt = 2−(1/(2α))ln(α^2 +α +1)−(3/(2α(√3))){ arctan(((2α +1)/(2(√3))))−arctan((1/(2(√3))))} ⇒ f(x) =((α−1)/α)ln(1−α) −1 +ln(α^2 +α +1)−2+(1/(2α))ln(α^2 +α +1) +((√3)/(2α)){ arctan(((2α +1)/(2(√3))))−arctan((1/(2(√3))))} with α =^3 (√x)](Q62029.png)
$${let}\:{use}\:{another}\:{way}\:\:{let}\:\alpha\:=^{\mathrm{3}} \sqrt{{x}}\:\Rightarrow \\ $$ $${f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−\left(\alpha{t}\right)^{\mathrm{3}} \right){dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left\{\left(\mathrm{1}−\alpha{t}\right)\left(\mathrm{1}+\alpha{t}\:+\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \right)\right\}{dt} \\ $$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−\alpha{t}\right){dt}\:+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}\right){dt} \\ $$ $${changement}\:\mathrm{1}−\alpha{t}\:={u}\:{give}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−\alpha{t}\right){dt}\:=\frac{−\mathrm{1}}{\alpha}\:\int_{\mathrm{1}} ^{\mathrm{1}−\alpha} \:{ln}\left({u}\right)\:{du} \\ $$ $$=−\frac{\mathrm{1}}{\alpha}\left[{ulnu}−{u}\right]_{\mathrm{1}} ^{\mathrm{1}−\alpha} \:=−\frac{\mathrm{1}}{\alpha}\left\{\:\left(\mathrm{1}−\alpha\right){ln}\left(\mathrm{1}−\alpha\right)−\left(\mathrm{1}−\alpha\right)\:+\mathrm{1}\right\} \\ $$ $$=−\frac{\mathrm{1}}{\alpha}\left\{\:\left(\mathrm{1}−\alpha\right){ln}\left(\mathrm{1}−\alpha\right)\:+\alpha\right\}\:=\frac{\alpha−\mathrm{1}}{\alpha}{ln}\left(\mathrm{1}−\alpha\right)−\mathrm{1}\:.\:{by}\:{parts}\: \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}\right){dt}\:=\left[{t}\:{ln}\left(\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} {t}\:\:\:\frac{\mathrm{2}\alpha^{\mathrm{2}} {t}\:+\alpha}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}}\:{dt} \\ $$ $$={ln}\left(\alpha^{\mathrm{2}} \:+\alpha\:+\mathrm{1}\right)\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}\alpha^{\mathrm{2}} {t}^{\mathrm{2}} +\alpha{t}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}}\:{dt} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}}\:{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}\left(\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}\right)−\mathrm{2}\alpha{t}−\mathrm{2}\:+\alpha{t}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}}\:{dt} \\ $$ $$=\mathrm{2}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\alpha{t}\:+\mathrm{2}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}}\:{dt}\:=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}\alpha}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}\alpha^{\mathrm{2}} {t}\:+\alpha\:+\mathrm{3}\alpha}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}}\:{dt} \\ $$ $$=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}\alpha}\left[{ln}\left(\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\frac{\mathrm{3}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}\:+\mathrm{1}}\:=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{{t}}{\alpha}\:+\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:++\frac{\mathrm{2}{t}}{\mathrm{2}\alpha}\:+\frac{\mathrm{1}}{\mathrm{4}\alpha^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:−\frac{\mathrm{1}}{\mathrm{4}\alpha^{\mathrm{2}} }} \\ $$ $$=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\left({t}\:+\frac{\mathrm{1}}{\mathrm{2}\alpha}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\alpha^{\mathrm{2}} }}\:=_{{t}+\frac{\mathrm{1}}{\mathrm{2}\alpha}=\frac{\sqrt{\mathrm{3}}}{\alpha}\:{x}} \:\:\:\:\:\:\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:\int_{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}} ^{\frac{\mathrm{2}\alpha+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}} \:\:\:\:\frac{\alpha^{\mathrm{2}} }{\mathrm{3}}\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{3}}}{\alpha}\:{dx}\:\:\left({we}\:{take}\:{x}>\mathrm{0}\right) \\ $$ $$=\frac{\mathrm{1}}{\alpha\sqrt{\mathrm{3}}}\:\left\{\:{arctan}\left(\frac{\mathrm{2}\alpha+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)\right\}\:\Rightarrow \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}}{\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \:+\alpha{t}+\mathrm{1}}\:{dt}\:=\:\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}\alpha}{ln}\left(\alpha^{\mathrm{2}} \:+\alpha\:+\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{2}\alpha\sqrt{\mathrm{3}}}\left\{\:{arctan}\left(\frac{\mathrm{2}\alpha\:+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)\right\}\:\Rightarrow \\ $$ $${f}\left({x}\right)\:=\frac{\alpha−\mathrm{1}}{\alpha}{ln}\left(\mathrm{1}−\alpha\right)\:−\mathrm{1}\:+{ln}\left(\alpha^{\mathrm{2}} \:+\alpha\:+\mathrm{1}\right)−\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}\alpha}{ln}\left(\alpha^{\mathrm{2}} \:+\alpha\:+\mathrm{1}\right) \\ $$ $$+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\alpha}\left\{\:{arctan}\left(\frac{\mathrm{2}\alpha\:+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)\right\}\:\:{with}\:\alpha\:=^{\mathrm{3}} \sqrt{{x}} \\ $$ $$ \\ $$
Commented bymaxmathsup by imad last updated on 14/Jun/19
$${f}\left({x}\right)\:=\frac{\left(\alpha−\mathrm{1}\right){ln}\left(\mathrm{1}−\alpha\right)}{\alpha}\:−\mathrm{3}\:\:+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\alpha}\right){ln}\left(\alpha^{\mathrm{2}} \:+\alpha\:+\mathrm{1}\right)\: \\ $$ $$+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\alpha}\left\{\:{arctan}\left(\frac{\mathrm{2}\alpha\:+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)\right\}\:. \\ $$
Commented bymaxmathsup by imad last updated on 14/Jun/19
$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{t}^{\mathrm{3}} \right)\:={f}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\right)\:\:{we}\:{have}\:\alpha\:={x}^{\frac{\mathrm{1}}{\mathrm{3}}} \:\Rightarrow\:\alpha\:=\left(\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:=\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{6}}} \:=\frac{\mathrm{1}}{\left(^{\mathrm{6}} \sqrt{\mathrm{2}}\right)}\:\Rightarrow \\ $$ $${f}\left({x}\right)\:=\frac{\left(\alpha−\mathrm{1}\right){ln}\left(\mathrm{1}−\alpha\right)}{\alpha}\:−\mathrm{3}\:+\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\alpha}\right){ln}\left(\alpha^{\mathrm{2}} \:+\alpha\:+\mathrm{1}\right) \\ $$ $$+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\alpha}\left\{\:{arctan}\left(\frac{\mathrm{2}\alpha\:+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)−{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)\right\}\:. \\ $$
Commented bymaxmathsup by imad last updated on 14/Jun/19
$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{ln}\left(\mathrm{1}−{sin}\theta\:{t}^{\mathrm{3}} \right){d}\theta\:={f}\left({sin}\theta\right)\:\:\:{we}\:\alpha\:=^{\mathrm{3}} \sqrt{{x}}\:\Rightarrow\alpha\:=^{\mathrm{3}} \sqrt{{sin}\theta} \\ $$ $${so}\:{the}\:{value}\:{of}\:{this}\:{integral}\:{is}\:{known}. \\ $$