1-1-1-x-3-dx-

Question Number 129057 by pipin last updated on 12/Jan/21

\int_{1}^{\infty} \frac{1}{1 + x^{3}} dx = \dots

Answered by MJS_new last updated on 12/Jan/21

\int \frac{d x}{x^{3} + 1} = \int \frac{d x}{(x + 1) (x^{2} - x + 1)} =

= \frac{1}{3} \int \frac{d x}{x + 1} - \frac{1}{3} \int \frac{x - 2}{x^{2} - x + 1} d x =

= \frac{1}{3} \int \frac{d x}{x + 1} - \frac{1}{6} \int \frac{2 x - 1}{x^{2} - x + 1} d x + \frac{1}{2} \int \frac{d x}{x^{2} - x + 1} =

= \frac{1}{3} \ln ∣ x + 1 ∣ - \frac{1}{6} \ln (x^{2} - x + 1) + \frac{\sqrt{3}}{3} \arctan \frac{\sqrt{3} (2 x - 1)}{3} + C

\Rightarrow answer is \frac{π \sqrt{3}}{9} - \frac{\ln 2}{3}

Answered by Dwaipayan Shikari last updated on 12/Jan/21

\int_{0}^{\infty} \frac{1}{1 + x^{3}} - \frac{1}{3} \int_{0}^{1} \frac{1}{1 + x} - \frac{x - 2}{x^{2} - x + 1} d x

= \frac{1}{3} \int_{0}^{\infty} \frac{u^{- \frac{2}{3}}}{1 + u} d u - \frac{1}{3} \int_{0}^{1} \frac{1}{1 + x} + \frac{1}{6} \int_{0}^{1} \frac{2 x - 1}{x^{2} - x + 1} - \frac{1}{2} \int \frac{1}{x^{2} - x + 1}

= \frac{1}{3} . \frac{π}{s i n (\frac{π}{3})} - \frac{1}{3} l o g (2) - \frac{1}{\sqrt{3}} {[{t a n}^{- 1} \frac{2 x - 1}{\sqrt{3}}]}_{0}^{1}

= \frac{2 π}{3 \sqrt{3}} - \frac{1}{3} l o g (2) - \frac{π}{3 \sqrt{3}} = \frac{π}{3 \sqrt{3}} - \frac{1}{3} l o g (2)

Commented by MJS_new last updated on 12/Jan/21

I had misread the lower border \dots

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