Question Number 45635 by maxmathsup by imad last updated on 14/Oct/18
$$\left.\mathrm{1}\right){find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{xt}^{\mathrm{3}} \right){dt}\:\:{with}\:\:\mid{x}\mid<\mathrm{1} \\ $$ $$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{2}+{t}^{\mathrm{3}} \right){dt}\:. \\ $$
Commented bymaxmathsup by imad last updated on 17/Oct/18
![1) changement^3 (√x)t =u give f(x)= ∫_0 ^3_(√x) ln(1+u^3 ) (du/((^3 (√x)))) =(1/((^3 (√x)))) ∫_0 ^3_(√x) ln(1+u^3 )du but we have proved that ∫ ln(1+x^3 )dx =xln(1+x^3 )−3x +ln∣x+1∣ −(1/2)ln(x^2 −x+1) +(√3)arctan(((2x−1)/(√3))) f(x)= (1/((^3 (√x))))[uln(1+u^3 )−3u +ln∣u+1∣−(1/2)ln(u^2 −u+1)+(√3)arctan(((2x−1)/(√3)))]_0 ^3_(√x) =(1/((^3 (√x)))){^3 (√x)ln(1+x)−3^ (^3 (√x))+ln∣1+^3 (√x)∣−(1/2)ln{ u^(2/3) −^3 (√x)+1} +(√3)arctan(((2(^3 (√x))−1)/(√3))) +((π(√3))/6)} .](Q45888.png)
$$\left.\mathrm{1}\right)\:{changement}\:^{\mathrm{3}} \sqrt{{x}}{t}\:\:={u}\:\:{give}\: \\ $$ $${f}\left({x}\right)=\:\int_{\mathrm{0}} ^{\mathrm{3}_{\sqrt{{x}}} } {ln}\left(\mathrm{1}+{u}^{\mathrm{3}} \right)\:\frac{{du}}{\left(^{\mathrm{3}} \sqrt{{x}}\right)}\:=\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{{x}}\right)}\:\int_{\mathrm{0}} ^{\mathrm{3}_{\sqrt{{x}}} } \:\:{ln}\left(\mathrm{1}+{u}^{\mathrm{3}} \right){du}\:{but}\:{we}\:{have}\:{proved}\:{that} \\ $$ $$\int\:{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx}\:={xln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\mathrm{3}{x}\:+{ln}\mid{x}+\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right) \\ $$ $${f}\left({x}\right)=\:\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{{x}}\right)}\left[{uln}\left(\mathrm{1}+{u}^{\mathrm{3}} \right)−\mathrm{3}{u}\:+{ln}\mid{u}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({u}^{\mathrm{2}} −{u}+\mathrm{1}\right)+\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\mathrm{3}_{\sqrt{{x}}} } \\ $$ $$=\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{{x}}\right)}\left\{^{\mathrm{3}} \sqrt{{x}}{ln}\left(\mathrm{1}+{x}\right)−\mathrm{3}^{} \left(^{\mathrm{3}} \sqrt{{x}}\right)+{ln}\mid\mathrm{1}+^{\mathrm{3}} \sqrt{{x}}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\:{u}^{\frac{\mathrm{2}}{\mathrm{3}}} −^{\mathrm{3}} \sqrt{{x}}+\mathrm{1}\right\}\right. \\ $$ $$\left.+\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{2}\left(^{\mathrm{3}} \sqrt{{x}}\right)−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\:+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{6}}\right\}\:. \\ $$ $$ \\ $$
Commented bymaxmathsup by imad last updated on 17/Oct/18
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{2}+{t}^{\mathrm{3}} \right){dt}\:\Rightarrow\:{A}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left\{\mathrm{2}\left(\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{3}} \right)\right\}{dt} \\ $$ $$={ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{3}} \right){dt} \\ $$ $$={ln}\left(\mathrm{2}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:={ln}\left(\mathrm{2}\right)\:+^{\mathrm{3}} \sqrt{\mathrm{2}}\left\{^{\mathrm{3}} \sqrt{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{3}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)\:+{ln}\mid\mathrm{1}+^{\mathrm{3}} \sqrt{\mathrm{2}}\mid\right. \\ $$ $$\left.−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\:\:\mathrm{2}^{−\frac{\mathrm{2}}{\mathrm{3}}} −\frac{\mathrm{1}}{\left(^{\mathrm{3}} \sqrt{{x}}\right)}\:+\mathrm{1}\right\}\:+\sqrt{\mathrm{3}}{arctan}\left(\frac{\frac{\mathrm{2}}{\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)}−\mathrm{1}}{\sqrt{\mathrm{3}}}\right)+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{6}}\right\}\:. \\ $$