ϝ = ∫ (dx/(x^(2n+1) (x^n −1)))


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Question Number 129646 by liberty last updated on 17/Jan/21

$ϝ = \int \frac{d x}{x^{2 n + 1} (x^{n} - 1)}$
	Answered by TheSupreme last updated on 17/Jan/21
	$x^n =t x=t^(1/n) x^(2n+1) =t^(2+(1/n)) dx=(1/n)t^((1/n)−1) dt ∫(((1/n)t^((1/n)−1) )/(t^(2+(1/n)) (t−1)))=∫(1/n) (1/(t^3 (t−1)))dt ((At^2 +Bt+C)/t^3 )+(D/(t−1))=(1/D) { (((t^3 ) A+D=0)),(((t^2 ) B−A=0)),(((t) −B+C=0)),(((1) −C=1)) :} F=(1/n)∫−((t^2 +t+1)/t^3 )+(1/(t−1)) dt F=(1/n)∫−(1/t)−(1/t^2 )−(1/t^3 )+(1/(t−1))dt F=(1/n)[−log(t)+(1/t)+(1/(2t^2 ))+log(t−1)+c] F=(1/n)log((x^n −1)/x^n ) +(1/x^n )+(1/(2x^(2n) ))+c$
	$x^{n} = t$ $x = t^{\frac{1}{n}}$ $x^{2 n + 1} = t^{2 + \frac{1}{n}}$ $d x = \frac{1}{n} t^{\frac{1}{n} - 1} d t$ $\int \frac{\frac{1}{n} t^{\frac{1}{n} - 1}}{t^{2 + \frac{1}{n}} (t - 1)} = \int \frac{1}{n} \frac{1}{t^{3} (t - 1)} d t$ $\frac{{A t}^{2} + B t + C}{t^{3}} + \frac{D}{t - 1} = \frac{1}{D}$ ${\begin{cases} (t^{3}) A + D = 0 \\ (t^{2}) B - A = 0 \\ (t) - B + C = 0 \\ (1) - C = 1 \end{cases}$ $F = \frac{1}{n} \int - \frac{t^{2} + t + 1}{t^{3}} + \frac{1}{t - 1} d t$ $F = \frac{1}{n} \int - \frac{1}{t} - \frac{1}{t^{2}} - \frac{1}{t^{3}} + \frac{1}{t - 1} d t$ $F = \frac{1}{n} [- l o g (t) + \frac{1}{t} + \frac{1}{2 t^{2}} + l o g (t - 1) + c]$ $F = \frac{1}{n} \log \frac{x^{n} - 1}{x^{n}} + \frac{1}{x^{n}} + \frac{1}{{2 x}^{2 n}} + c$
	Answered by liberty last updated on 17/Jan/21

	$ϝ = - \frac{1}{2 n} \int \frac{1}{x^{n} - 1} d (\frac{1}{x^{2 n}})$ $ϝ = - \frac{1}{2 n} \int \frac{\frac{1}{x^{n}}}{1 - \frac{1}{x^{n}}} d (\frac{1}{x^{2 n}})$ $let \frac{1}{x^{n}} = r \Rightarrow r^{2} = \frac{1}{x^{2 n}} \land 2 r d r = d (\frac{1}{x^{2 n}})$ $ϝ = - \frac{1}{2 n} \int \frac{r}{1 - r} (2 r d r)$ $ϝ = - \frac{1}{n} \int \frac{1 - (1 - r^{2})}{1 - r} d r$ $ϝ = \frac{1}{n} [\ln ∣ 1 - r ∣ + \int (1 + r) d r]$ $ϝ = \frac{1}{n} [\ln ∣ 1 - \frac{1}{x^{n}} ∣ + \frac{1}{x^{n}} + \frac{1}{{2 x}^{2 n}}] + C$
	Answered by mathmax by abdo last updated on 17/Jan/21

	$I = \int \frac{dx}{x^{2 n + 1} (x^{n} - 1)} changement x = t^{\frac{1}{n}} give$ $I = \frac{1}{n} \int \frac{t^{\frac{1}{n} - 1}}{t^{\frac{2 n + 1}{n}} (t - 1)} dt = \frac{1}{n} \int \frac{t^{\frac{1}{n} - 1} . t^{- 2 - \frac{1}{n}}}{t - 1} dt$ $= \frac{1}{n} \int \frac{dt}{t^{3} (t - 1)} let decompose F (t) = \frac{1}{t^{3} (t - 1)}$ $F (t) = \frac{a}{t} + \frac{b}{t^{2}} + \frac{c}{t^{3}} + \frac{d}{t - 1}$ $c = - 1, d = 1 \Rightarrow F (t) = \frac{a}{t} + \frac{b}{t^{2}} - \frac{1}{t^{3}} + \frac{1}{t - 1}$ $\lim_{t \to + \infty} tF (t) = 0 = a + 1 \Rightarrow a = - 1 \Rightarrow F (t) = - \frac{1}{t} + \frac{b}{t^{2}} - \frac{1}{t^{3}} + \frac{1}{t - 1}$ $F (- 1) = \frac{1}{2} = 1 + b + 1 - \frac{1}{2} = b + 2 - \frac{1}{2} = b + \frac{3}{2} \Rightarrow b = \frac{1}{2} - \frac{3}{2} = - 1 \Rightarrow$ $F (t) = - \frac{1}{t} - \frac{1}{t^{2}} - \frac{1}{t^{3}} + \frac{1}{t - 1} \Rightarrow \int F (t) dt = - \ln ∣ t ∣ + \frac{1}{t} + \frac{1}{{2 t}^{2}} + \ln ∣ t - 1 ∣ + C$ $= \ln ∣ \frac{t - 1}{t} ∣ + \frac{1}{t} + \frac{1}{{2 t}^{2}} + C = \ln ∣ \frac{x^{n} - 1}{x^{n}} ∣ + \frac{1}{x^{n}} + \frac{1}{{2 x}^{2 n}} + C$
	Commented by mathmax by abdo last updated on 17/Jan/21

	$\Rightarrow I = \frac{1}{n} \ln ∣ \frac{x^{n} - 1}{x^{n}} ∣ + \frac{1}{{nx}^{n}} + \frac{1}{{2 nx}^{2 n}} + C$
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