Question Number 205371 by 073 last updated on 19/Mar/24
Commented by lepuissantcedricjunior last updated on 19/Mar/24
![I=∫_0 ^2 log(x^3 +8)dx posons { ((u=log(x^3 +8))),((v′=1)) :}⇔ { ((u′=((3x^2 )/((x^3 +8)ln10)))),((v=x)) :} I=[xlog(x^3 +8)]_0 ^2 −(3/(ln10))∫_0 ^2 (1−(8/(x^3 +8)))dx =2log(16)−(6/(ln10))+((24)/(ln10))∫_0 ^2 ((a/(x+2))+((bx+c)/(x^2 −2x+4)))dx { ((a+b=0=>a=−b)),((−2a+2b+c=0=>−4a+c=0=>c=4a)) :}4a+2c=1 4a+2c=1=>8a=1=>a=(1/8) b=−(1/8);c=(1/2) ⇒I=8log(2)−(6/(ln10))+((24)/(ln10))∫_0 ^2 (((1/8)/x)−(((1/8)x−(1/2))/(x^2 −2x+4)))dx =8log(2)−(6/(ln10))+(3/(ln10))∫_0 ^2 ((1/x)−(1/2)(((2x−2−6)/(x^2 −2x+4))))dx =8log(2)−(6/(ln10))+(3/(ln10))ln2−((3ln4)/(2ln10))+(9/(ln10))∫_0 ^2 (dx/(3[1+(((x−1)/( (√3))))^2 ])) =8log(2)−(6/(ln10))+((3(√3))/(ln10))[arctan(((x−1)/( (√3))))]_0 ^2 =8log(2)−(6/(ln10))+((3(√3))/(ln10))((𝛑/6)−(−(𝛑/6))) =8log(2)−(1/(ln10))(6−𝛑(√3)) ============================================= ..................le puissant Dr.............................. =========================================================](https://www.tinkutara.com/question/Q205376.png)
$$\boldsymbol{\mathrm{I}}=\int_{\mathrm{0}} ^{\mathrm{2}} \boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{8}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\boldsymbol{\mathrm{posons}}\:\begin{cases}{\boldsymbol{\mathrm{u}}=\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{8}\right)}\\{\boldsymbol{\mathrm{v}}'=\mathrm{1}}\end{cases}\Leftrightarrow\begin{cases}{\boldsymbol{\mathrm{u}}'=\frac{\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} }{\left(\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{8}\right)\boldsymbol{\mathrm{ln}}\mathrm{10}}}\\{\boldsymbol{\mathrm{v}}=\boldsymbol{\mathrm{x}}}\end{cases} \\ $$$$\boldsymbol{\mathrm{I}}=\left[\boldsymbol{\mathrm{xlog}}\left(\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{8}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} −\frac{\mathrm{3}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{8}}{\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{8}}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\:\:=\mathrm{2}\boldsymbol{\mathrm{log}}\left(\mathrm{16}\right)−\frac{\mathrm{6}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}+\frac{\mathrm{24}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{x}}+\mathrm{2}}+\frac{\boldsymbol{\mathrm{bx}}+\boldsymbol{\mathrm{c}}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{4}}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}=\mathrm{0}=>\boldsymbol{\mathrm{a}}=−\boldsymbol{\mathrm{b}}}\\{−\mathrm{2}\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}=\mathrm{0}=>−\mathrm{4}\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}=\mathrm{0}=>\boldsymbol{\mathrm{c}}=\mathrm{4}\boldsymbol{\mathrm{a}}}\end{cases}\mathrm{4}\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{c}}=\mathrm{1} \\ $$$$\mathrm{4}\boldsymbol{\mathrm{a}}+\mathrm{2}\boldsymbol{\mathrm{c}}=\mathrm{1}=>\mathrm{8}\boldsymbol{\mathrm{a}}=\mathrm{1}=>\boldsymbol{\mathrm{a}}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\boldsymbol{\mathrm{b}}=−\frac{\mathrm{1}}{\mathrm{8}};\boldsymbol{{c}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\boldsymbol{\mathrm{I}}=\mathrm{8}\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)−\frac{\mathrm{6}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}+\frac{\mathrm{24}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\frac{\frac{\mathrm{1}}{\mathrm{8}}}{\boldsymbol{\mathrm{x}}}−\frac{\frac{\mathrm{1}}{\mathrm{8}}\boldsymbol{\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{2}}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{4}}\right)\boldsymbol{\mathrm{dx}} \\ $$$$\:\:\:\:\:\:\:=\mathrm{8}\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)−\frac{\mathrm{6}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}+\frac{\mathrm{3}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}\boldsymbol{\mathrm{x}}−\mathrm{2}−\mathrm{6}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{4}}\right)\right)\boldsymbol{\mathrm{dx}} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{8}\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)−\frac{\mathrm{6}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}+\frac{\mathrm{3}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}\boldsymbol{\mathrm{ln}}\mathrm{2}−\frac{\mathrm{3}\boldsymbol{\mathrm{ln}}\mathrm{4}}{\mathrm{2}\boldsymbol{\mathrm{ln}}\mathrm{10}}+\frac{\mathrm{9}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\boldsymbol{\mathrm{dx}}}{\mathrm{3}\left[\mathrm{1}+\left(\frac{\boldsymbol{\mathrm{x}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \right]} \\ $$$$=\mathrm{8}\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)−\frac{\mathrm{6}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}\left[\boldsymbol{\mathrm{arctan}}\left(\frac{\boldsymbol{\mathrm{x}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\mathrm{8}\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)−\frac{\mathrm{6}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}\left(\frac{\boldsymbol{\pi}}{\mathrm{6}}−\left(−\frac{\boldsymbol{\pi}}{\mathrm{6}}\right)\right) \\ $$$$=\mathrm{8}\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\boldsymbol{\mathrm{ln}}\mathrm{10}}\left(\mathrm{6}−\boldsymbol{\pi}\sqrt{\mathrm{3}}\right) \\ $$$$============================================= \\ $$$$………………{le}\:{puissant}\:\boldsymbol{{D}}{r}………………………… \\ $$$$========================================================= \\ $$$$ \\ $$$$ \\ $$
Commented by 073 last updated on 19/Mar/24
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Answered by mathzup last updated on 21/Mar/24
![x=2t ⇒I=2∫_0 ^1 log(8t^3 +8)dt =2∫_0 ^1 (3ln2 +ln(1+t^3 ))dt =6ln2 +2∫_0 ^1 ln(1+t^3 )dt but ∫_0 ^1 ln(1+t^3 )dt=[tln(1+t^3 )]_0 ^1 −∫_0 ^1 ((t(3t^2 ))/(1+t^3 ))dt =ln2−3∫_0 ^1 ((t^3 +1−1)/(1+t^3 ))dt=ln2−3+3∫_0 ^1 (dt/(1+t^3 )) ∫_0 ^1 (dt/(1+t^3 ))=∫_0 ^1 Σ(−1)^n t^(3n) dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^(3n) dt=Σ_(n=0) ^∞ (((−1)^n )/(3n+1)) =Σ_(n=0) ^∞ (1/(6n+1))−Σ_(n=0) (1/(3(2n+1)+1)) =Σ_(n=0) ^∞ ((1/(6n+1))−(1/(6n+4))) =3Σ_(n=0) ^∞ (1/((6n+1)(6n+4))) =(1/(12))Σ_(n=0) ^∞ (1/((n+(1/6))(n+(2/3)))) =(1/(12))(Ψ((2/3))−Ψ((1/6))×(1/((2/3)−(1/6))) =(1/6)(Ψ((2/3))−Ψ((1/6))) ⇒ I=ln2−3 +(1/2){Ψ((2/3))−Ψ((1/6))}](https://www.tinkutara.com/question/Q205425.png)
$${x}=\mathrm{2}{t}\:\Rightarrow{I}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\mathrm{8}{t}^{\mathrm{3}} +\mathrm{8}\right){dt} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{3}{ln}\mathrm{2}\:+{ln}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)\right){dt} \\ $$$$=\mathrm{6}{ln}\mathrm{2}\:+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}^{\mathrm{3}} \right){dt} \\ $$$${but}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{t}^{\mathrm{3}} \right){dt}=\left[{tln}\left(\mathrm{1}+{t}^{\mathrm{3}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}\left(\mathrm{3}{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{3}} }{dt} \\ $$$$={ln}\mathrm{2}−\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{3}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{3}} }{dt}={ln}\mathrm{2}−\mathrm{3}+\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{3}} }=\int_{\mathrm{0}} ^{\mathrm{1}} \Sigma\left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{3}{n}} {dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{3}{n}} {dt}=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{3}{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{6}{n}+\mathrm{1}}−\sum_{{n}=\mathrm{0}} \:\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{2}{n}+\mathrm{1}\right)+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{6}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{6}{n}+\mathrm{4}}\right) \\ $$$$=\mathrm{3}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{6}{n}+\mathrm{1}\right)\left(\mathrm{6}{n}+\mathrm{4}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({n}+\frac{\mathrm{1}}{\mathrm{6}}\right)\left({n}+\frac{\mathrm{2}}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\Psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right)×\frac{\mathrm{1}}{\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{6}}}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left(\Psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\right)\:\Rightarrow \\ $$$${I}={ln}\mathrm{2}−\mathrm{3}\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\Psi\left(\frac{\mathrm{2}}{\mathrm{3}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\right\} \\ $$