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Question Number 203385 by patrice last updated on 18/Jan/24

	Answered by Mathspace last updated on 18/Jan/24
	$I=∫_0 ^1 (dx/(1+x^3 )) ⇒I=∫_0 ^1 Σ_(n=0) ^∞ (−1)^n x^(3n) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(3n) dx =Σ_(n=0) ^∞ (((−1)^n )/(3n+1)) =Σ_(n=0) ^∞ (1/(6n+1))−Σ_(n=0) ^∞ (1/(3(2n+1)+1)) =Σ_(n=0) ^∞ ((1/(6n+1))−(1/(6n+4))) =Σ_(n=0) ^∞ (3/((6n+1)(6n+4))) =(1/(12))Σ_(n=0) ^∞ (1/((n+(1/6))(n+(4/6)))) =(1/(12)){((Ψ((4/6))−Ψ((1/6)))/((4/6)−(1/6)))} =(1/(12))×2(Ψ((4/6))−Ψ((1/6))) =(1/6)(Ψ((2/3))−Ψ((1/6))) now use Ψ(s)=−γ+∫_0 ^1 ((1−x^(s−1) )/(1−x))dx to calculate Ψ((2/3))and Ψ((1/6))$
	$I = \int_{0}^{1} \frac{d x}{1 + x^{3}} \Rightarrow I = \int_{0}^{1} \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{3 n} d x$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{1} x^{3 n} d x$ $= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{3 n + 1}$ $= \sum_{n = 0}^{\infty} \frac{1}{6 n + 1} - \sum_{n = 0}^{\infty} \frac{1}{3 (2 n + 1) + 1}$ $= \sum_{n = 0}^{\infty} (\frac{1}{6 n + 1} - \frac{1}{6 n + 4})$ $= \sum_{n = 0}^{\infty} \frac{3}{(6 n + 1) (6 n + 4)}$ $= \frac{1}{12} \sum_{n = 0}^{\infty} \frac{1}{(n + \frac{1}{6}) (n + \frac{4}{6})}$ $= \frac{1}{12} {\frac{Ψ (\frac{4}{6}) - Ψ (\frac{1}{6})}{\frac{4}{6} - \frac{1}{6}}}$ $= \frac{1}{12} \times 2 (Ψ (\frac{4}{6}) - Ψ (\frac{1}{6}))$ $= \frac{1}{6} (Ψ (\frac{2}{3}) - Ψ (\frac{1}{6}))$ $n o w u s e Ψ (s) = - γ + \int_{0}^{1} \frac{1 - x^{s - 1}}{1 - x} d x$ $t o c a l c u l a t e Ψ (\frac{2}{3}) a n d Ψ (\frac{1}{6})$
	Commented by MathematicalUser2357 last updated on 04/Mar/24
	$(1/6)[(−γ+∫_0 ^1 ((1−(1/( (x)^(1/3) )))/(1−x))dx)−{−γ+∫_0 ^1 ((1−(1/( ((x)^(1/6) )^5 )))/(1−x))dx}]$
	$\frac{1}{6} [(- γ + \int_{0}^{1} \frac{1 - \frac{1}{\sqrt[3]{x}}}{1 - x} d x) - {- γ + \int_{0}^{1} \frac{1 - \frac{1}{{(\sqrt[6]{x})}^{5}}}{1 - x} d x}]$
	Answered by Mathspace last updated on 18/Jan/24

	$c l a s s i c m e t h o d$ $f (x) = \frac{1}{x^{3} + 1} = \frac{1}{(x + 1) (x^{2} - x + 1)}$ $= \frac{a}{x + 1} + \frac{b x + c}{x^{2} - x + 1}$ $a = \frac{1}{3}$ ${l i m}_{x \to + \infty} x f (x) = 0 = a + b \Rightarrow$ $b = - \frac{1}{3}$ $f (0) = 1 = a + c \Rightarrow c = 1 - a$ $= 1 - \frac{1}{3} = \frac{2}{3} \Rightarrow$ $f (x) = \frac{1}{3 (x + 1)} + \frac{- \frac{1}{3} x + \frac{2}{3}}{x^{2} - x + 1}$ $\Rightarrow \int f (x) d x = \int \frac{d x}{3 (x + 1)}$ $- \frac{1}{3} \int \frac{x - 2}{x^{2} - x + 1} d x$ $= \frac{1}{3} l n ∣ x + 1 ∣ - \frac{1}{6} \int \frac{2 x - 1 - 3}{x^{2} - x + 1} d x$ $= \frac{1}{3} l n ∣ x + 1 ∣ - \frac{1}{6} l n (x^{2} - x + 1)$ $+ \frac{1}{2} \int \frac{d x}{x^{2} - x + 1} a n d$ $\int \frac{d x}{x^{2} - x + 1} = \int \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}$ $=_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} t} \frac{\sqrt{3}}{2} \int \frac{d t}{\frac{3}{4} (1 + t^{2})}$ $= \frac{4}{3} . \frac{\sqrt{3}}{2} \int \frac{d t}{1 + t^{2}} = \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}})$ $\Rightarrow \int f (x) d x = \frac{1}{3} l n ∣ x + 1 ∣$ $- \frac{1}{6} l n (x^{2} - x + 1) + \frac{1}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}})$ $I = \int_{0}^{1} f (x) d x$ $= [\frac{1}{3} l n ∣ x + 1 ∣ - \frac{1}{6} l n (x^{2} - x + 1)$ ${+ \frac{1}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}})]}_{0}^{1}$ $= \frac{1}{3} l n (2) + \frac{1}{\sqrt{3}} a r c t a n (\frac{1}{\sqrt{3}})$ $+ \frac{1}{\sqrt{3}} a r c t a n (\frac{1}{\sqrt{3}})$ $= \frac{l n 2}{3} + \frac{2}{\sqrt{3}} \times \frac{π}{3} = \frac{l n 2}{2} + \frac{2 π}{3 \sqrt{3}}$
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