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

Question Number 128953 by mathmax by abdo last updated on 11/Jan/21

calculate \int_{0}^{1} \ln (1 + x^{6}) dx

Answered by Lordose last updated on 12/Jan/21

Ω = \int_{0}^{1} \ln (1 + x^{6}) dx = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} x^{6 n} dx

Ω = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n (6 n + 1)} = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{n} - 6 \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{6 n + 1}

Ω = \log (2) + (π - 6 + 2 \sqrt{3} \coth^{- 1} (\sqrt{3}))

Ω \approx 0.1157 . .

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