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calculate-0-1-ln-1-x-6-dx-




Question Number 128953 by mathmax by abdo last updated on 11/Jan/21
calculate ∫_0 ^1 ln(1+x^6 )dx
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{6}} \right)\mathrm{dx} \\ $$
Answered by Lordose last updated on 12/Jan/21
  Ω = ∫_0 ^( 1) ln(1+x^6 )dx = Σ_(n=1) ^∞ (((−1)^(n−1) )/n)∫_0 ^( 1) x^(6n) dx  Ω = Σ_(n=1) ^∞  (((−1)^(n−1) )/(n(6n+1))) = Σ_(n=1) ^∞ (((−1)^(n−1) )/n) −  6Σ_(n=1) ^∞ (((−1)^(n−1) )/(6n+1))  Ω = log(2) + (π− 6 + 2(√3)coth^(−1) ((√3)))  Ω ≈ 0.1157..
$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{x}^{\mathrm{6}} \right)\mathrm{dx}\:=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{x}^{\mathrm{6n}} \mathrm{dx} \\ $$$$\Omega\:=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}\left(\mathrm{6n}+\mathrm{1}\right)}\:=\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}}\:−\:\:\mathrm{6}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{6n}+\mathrm{1}} \\ $$$$\Omega\:=\:\mathrm{log}\left(\mathrm{2}\right)\:+\:\left(\pi−\:\mathrm{6}\:+\:\mathrm{2}\sqrt{\mathrm{3}}\mathrm{coth}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}\right)\right) \\ $$$$\Omega\:\approx\:\mathrm{0}.\mathrm{1157}.. \\ $$

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