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Question Number 205371 by 073 last updated on 19/Mar/24

Commented by lepuissantcedricjunior last updated on 19/Mar/24
$I=∫_0 ^2 log(x^3 +8)dx posons { ((u=log(x^3 +8))),((v′=1)) :}⇔ { ((u′=((3x^2 )/((x^3 +8)ln10)))),((v=x)) :} I=[xlog(x^3 +8)]_0 ^2 −(3/(ln10))∫_0 ^2 (1−(8/(x^3 +8)))dx =2log(16)−(6/(ln10))+((24)/(ln10))∫_0 ^2 ((a/(x+2))+((bx+c)/(x^2 −2x+4)))dx { ((a+b=0=>a=−b)),((−2a+2b+c=0=>−4a+c=0=>c=4a)) :}4a+2c=1 4a+2c=1=>8a=1=>a=(1/8) b=−(1/8);c=(1/2) ⇒I=8log(2)−(6/(ln10))+((24)/(ln10))∫_0 ^2 (((1/8)/x)−(((1/8)x−(1/2))/(x^2 −2x+4)))dx =8log(2)−(6/(ln10))+(3/(ln10))∫_0 ^2 ((1/x)−(1/2)(((2x−2−6)/(x^2 −2x+4))))dx =8log(2)−(6/(ln10))+(3/(ln10))ln2−((3ln4)/(2ln10))+(9/(ln10))∫_0 ^2 (dx/(3[1+(((x−1)/( (√3))))^2 ])) =8log(2)−(6/(ln10))+((3(√3))/(ln10))[arctan(((x−1)/( (√3))))]_0 ^2 =8log(2)−(6/(ln10))+((3(√3))/(ln10))((𝛑/6)−(−(𝛑/6))) =8log(2)−(1/(ln10))(6−𝛑(√3)) ============================================= ..................le puissant Dr.............................. =========================================================$
$I = \int_{0}^{2} \log (x^{3} + 8) dx$ $posons {\begin{cases} u = \log (x^{3} + 8) \\ v^{'} = 1 \end{cases} \Leftrightarrow {\begin{cases} u^{'} = \frac{3 x^{2}}{(x^{3} + 8) \ln 10} \\ v = x \end{cases}$ $I = {[xlog (x^{3} + 8)]}_{0}^{2} - \frac{3}{\ln 10} \int_{0}^{2} (1 - \frac{8}{x^{3} + 8}) dx$ $= 2 \log (16) - \frac{6}{\ln 10} + \frac{24}{\ln 10} \int_{0}^{2} (\frac{a}{x + 2} + \frac{bx + c}{x^{2} - 2 x + 4}) dx$ ${\begin{cases} a + b = 0 => a = - b \\ - 2 a + 2 b + c = 0 => - 4 a + c = 0 => c = 4 a \end{cases} 4 a + 2 c = 1$ $4 a + 2 c = 1 => 8 a = 1 => a = \frac{1}{8}$ $b = - \frac{1}{8}; c = \frac{1}{2}$ $\Rightarrow I = 8 \log (2) - \frac{6}{\ln 10} + \frac{24}{\ln 10} \int_{0}^{2} (\frac{\frac{1}{8}}{x} - \frac{\frac{1}{8} x - \frac{1}{2}}{x^{2} - 2 x + 4}) dx$ $= 8 \log (2) - \frac{6}{\ln 10} + \frac{3}{\ln 10} \int_{0}^{2} (\frac{1}{x} - \frac{1}{2} (\frac{2 x - 2 - 6}{x^{2} - 2 x + 4})) dx$ $= 8 \log (2) - \frac{6}{\ln 10} + \frac{3}{\ln 10} \ln 2 - \frac{3 \ln 4}{2 \ln 10} + \frac{9}{\ln 10} \int_{0}^{2} \frac{dx}{3 [1 + {(\frac{x - 1}{\sqrt{3}})}^{2}]}$ $= 8 \log (2) - \frac{6}{\ln 10} + \frac{3 \sqrt{3}}{\ln 10} {[\arctan (\frac{x - 1}{\sqrt{3}})]}_{0}^{2}$ $= 8 \log (2) - \frac{6}{\ln 10} + \frac{3 \sqrt{3}}{\ln 10} (\frac{π}{6} - (- \frac{π}{6}))$ $= 8 \log (2) - \frac{1}{\ln 10} (6 - π \sqrt{3})$ $=============================================$ $. . . . . . . . . . . . . . . . . . l e p u i s s a n t D r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$ $=========================================================$
Commented by 073 last updated on 19/Mar/24

$? ? ? ? ? ? ?$
	Answered by mathzup last updated on 21/Mar/24
	$x=2t ⇒I=2∫_0 ^1 log(8t^3 +8)dt =2∫_0 ^1 (3ln2 +ln(1+t^3 ))dt =6ln2 +2∫_0 ^1 ln(1+t^3 )dt but ∫_0 ^1 ln(1+t^3 )dt=[tln(1+t^3 )]_0 ^1 −∫_0 ^1 ((t(3t^2 ))/(1+t^3 ))dt =ln2−3∫_0 ^1 ((t^3 +1−1)/(1+t^3 ))dt=ln2−3+3∫_0 ^1 (dt/(1+t^3 )) ∫_0 ^1 (dt/(1+t^3 ))=∫_0 ^1 Σ(−1)^n t^(3n) dt =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 t^(3n) dt=Σ_(n=0) ^∞ (((−1)^n )/(3n+1)) =Σ_(n=0) ^∞ (1/(6n+1))−Σ_(n=0) (1/(3(2n+1)+1)) =Σ_(n=0) ^∞ ((1/(6n+1))−(1/(6n+4))) =3Σ_(n=0) ^∞ (1/((6n+1)(6n+4))) =(1/(12))Σ_(n=0) ^∞ (1/((n+(1/6))(n+(2/3)))) =(1/(12))(Ψ((2/3))−Ψ((1/6))×(1/((2/3)−(1/6))) =(1/6)(Ψ((2/3))−Ψ((1/6))) ⇒ I=ln2−3 +(1/2){Ψ((2/3))−Ψ((1/6))}$
	$x = 2 t \Rightarrow I = 2 \int_{0}^{1} l o g (8 t^{3} + 8) d t$ $= 2 \int_{0}^{1} (3 l n 2 + l n (1 + t^{3})) d t$ $= 6 l n 2 + 2 \int_{0}^{1} l n (1 + t^{3}) d t$ $b u t \int_{0}^{1} l n (1 + t^{3}) d t = {[t l n (1 + t^{3})]}_{0}^{1} - \int_{0}^{1} \frac{t (3 t^{2})}{1 + t^{3}} d t$ $= l n 2 - 3 \int_{0}^{1} \frac{t^{3} + 1 - 1}{1 + t^{3}} d t = l n 2 - 3 + 3 \int_{0}^{1} \frac{d t}{1 + t^{3}}$ $\int_{0}^{1} \frac{d t}{1 + t^{3}} = \int_{0}^{1} Σ {(- 1)}^{n} t^{3 n} d t$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} t^{3 n} d t = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1}$ $= \sum_{n = 0}^{\infty} \frac{1}{6 n + 1} - \sum_{n = 0} \frac{1}{3 (2 n + 1) + 1}$ $= \sum_{n = 0}^{\infty} (\frac{1}{6 n + 1} - \frac{1}{6 n + 4})$ $= 3 \sum_{n = 0}^{\infty} \frac{1}{(6 n + 1) (6 n + 4)}$ $= \frac{1}{12} \sum_{n = 0}^{\infty} \frac{1}{(n + \frac{1}{6}) (n + \frac{2}{3})}$ $= \frac{1}{12} (Ψ (\frac{2}{3}) - Ψ (\frac{1}{6}) \times \frac{1}{\frac{2}{3} - \frac{1}{6}}$ $= \frac{1}{6} (Ψ (\frac{2}{3}) - Ψ (\frac{1}{6})) \Rightarrow$ $I = l n 2 - 3 + \frac{1}{2} {Ψ (\frac{2}{3}) - Ψ (\frac{1}{6})}$
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