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Question Number 45632 by maxmathsup by imad last updated on 14/Oct/18
1)find ∫ ln(1+x^3 )dx 2) calculate ∫_0 ^1 ln(1+x^3 )ex
$$\left.\mathrm{1}\right){find}\:\int\:{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right){ex} \\ $$
Commented by maxmathsup by imad last updated on 16/Oct/18
let I = ∫ ln(1+x^3 )dx by parts I = x ln(1+x^3 )− ∫ x ((3x^2 )/(1+x^3 ))dx =xln(1+x^3 )−3 ∫ ((1+x^3 −1)/(1+x^3 ))dx =xln(1+x^3 )−3x +3 ∫ (dx/(1+x^3 )) let decompose F(x)=(1/(1+x^3 )) F(x)=(1/((x+1)(x^2 −x+1))) =(a/(x+1)) +((bx +c)/(x^2 −x+1)) a =lim_(x→−1) (x+1)F(x)=(1/3) lim_(x→+∞) xF(x)=0 =a+b ⇒b=−a ⇒F(x)=(1/(3(x+1))) +((−(1/3)x +c)/(x^2 −x +1)) F(0) =1 = (1/3) +c ⇒c=(2/3) ⇒F(x)=(1/(3(x+1))) +((−(1/3)x +(2/3))/(x^2 −x +1)) ⇒3F(x) =(1/(x+1)) +((−x+2)/(x^2 −x+1)) ⇒∫ 3F(x)dx=ln∣x+1∣−(1/2)∫ ((2x−1−3)/(x^2 −x+1)) =ln∣x+1∣−(1/2)ln∣x^2 −x +1∣+(3/2)∫ (dx/(x^2 −x +1)) but ∫ (dx/(x^2 −x +1)) =∫ (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)u) (4/3) ∫ (1/(u^2 +1)) ((√3)/2)du =(2/( (√3))) ∫ (du/(1+u^2 )) =(2/( (√3))) arctan(((2x−1)/( (√3)))) ⇒∫ 3F(x)dx =ln∣x+1∣−(1/2)ln(x^2 −x+1) +(3/2) (2/( (√3)))arctan(((2x−1)/( (√3)))) ⇒ ∫ (dx/(1+x^3 )) =xln(1+x^3 ) −3x +ln∣x+1∣−(1/2)ln(x^2 −x+1) +(√3)arctan(((2x−1)/( (√3)))) +c
$${let}\:{I}\:=\:\int\:{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx}\:\:{by}\:{parts}\: \\ $$$${I}\:=\:{x}\:{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\:\int\:{x}\:\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{3}} }{dx}\:={xln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\mathrm{3}\:\int\:\:\frac{\mathrm{1}+{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} }{dx} \\ $$$$={xln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\mathrm{3}{x}\:\:+\mathrm{3}\:\int\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }\:\:{let}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{3}} } \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\:=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{bx}\:+{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}=−{a}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{3}}{x}\:+{c}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:+{c}\:\Rightarrow{c}=\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{3}}{x}\:+\frac{\mathrm{2}}{\mathrm{3}}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\ $$$$\Rightarrow\mathrm{3}{F}\left({x}\right)\:=\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+\frac{−{x}+\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\Rightarrow\int\:\mathrm{3}{F}\left({x}\right){dx}={ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{2}{x}−\mathrm{1}−\mathrm{3}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$={ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}^{\mathrm{2}} −{x}\:+\mathrm{1}\mid+\frac{\mathrm{3}}{\mathrm{2}}\int\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}\:{but} \\ $$$$\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}}\:=\int\:\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{u}} \:\frac{\mathrm{4}}{\mathrm{3}}\:\:\int\:\:\:\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{du} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\int\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\Rightarrow\int\:\mathrm{3}{F}\left({x}\right){dx} \\ $$$$={ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\frac{\mathrm{3}}{\mathrm{2}}\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:\Rightarrow \\ $$$$\int\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }\:={xln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\:−\mathrm{3}{x}\:\:+{ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+{c} \\ $$
Commented by maxmathsup by imad last updated on 16/Oct/18
∫_0 ^1 (dx/(1+x^3 )) =[xln(1+x^3 )−3x +ln((∣x+1∣)/( (√(x^2 −x+1)))) +(√3)arctan(((2x−1)/( (√3))))]_0 ^1 =ln(2)−3 +ln(2) +(√3) arctan((1/( (√3)))) +(√3)arctan((1/( (√3)))) =2ln(2)−3 +2 (√3)(π/6) =2ln(2) +((π(√3))/3) −3 .
$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }\:=\left[{xln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)−\mathrm{3}{x}\:+{ln}\frac{\mid{x}+\mathrm{1}\mid}{\:\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}\:+\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$={ln}\left(\mathrm{2}\right)−\mathrm{3}\:+{ln}\left(\mathrm{2}\right)\:+\sqrt{\mathrm{3}}\:{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+\sqrt{\mathrm{3}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$=\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{3}\:+\mathrm{2}\:\sqrt{\mathrm{3}}\frac{\pi}{\mathrm{6}}\:=\mathrm{2}{ln}\left(\mathrm{2}\right)\:+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{3}}\:−\mathrm{3}\:. \\ $$

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