∫_0 ^( 1) (x/(1+x^8 )) dx =?


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Question Number 137123 by bobhans last updated on 30/Mar/21

$\int_{0}^{1} \frac{x}{1 + x^{8}} dx = ?$
Commented by Ar Brandon last updated on 30/Mar/21
You're right, Sir. Greetings to you ! It's been quite a longtime since we last interracted. Haha !
Commented by MJS_new last updated on 30/Mar/21

$I get \frac{\sqrt{2}}{16} (π + 2 \ln (1 + \sqrt{2}))$
Commented by bobhans last updated on 30/Mar/21

$I got \frac{1}{8 \sqrt{2}} (\ln (\frac{2 + \sqrt{2}}{2 - \sqrt{2}}) + π)$
Commented by MJS_new last updated on 30/Mar/21

$\frac{1}{8 \sqrt{2}} = \frac{\sqrt{2}}{16}$ $\frac{2 + \sqrt{2}}{2 - \sqrt{2}} = \frac{{(2 + \sqrt{2})}^{2}}{(2 - \sqrt{2}) (2 + \sqrt{2})} = \frac{4 + 4 \sqrt{2} + 2}{4 - 2} = 3 + 2 \sqrt{2} = {(1 + \sqrt{2})}^{2}$ $\frac{1}{8 \sqrt{2}} (\ln \frac{2 + \sqrt{2}}{2 - \sqrt{2}} + π) = \frac{\sqrt{2}}{16} (π + 2 \ln (1 + \sqrt{2}))$
	Answered by Ar Brandon last updated on 30/Mar/21

	$I = \int_{0}^{1} \frac{x}{1 + x^{8}} dx \overset{u = x^{8}}{=} \frac{1}{8} \int_{0}^{1} \frac{u^{- \frac{6}{8}}}{1 + u} du = \frac{1}{8} \int_{0}^{1} \frac{u^{- \frac{3}{4}} (1 - u)}{1 - u^{2}} du$ $\overset{v = u^{2}}{=} \frac{1}{16} \int_{0}^{1} \frac{v^{- \frac{7}{8}} (1 - v^{\frac{1}{2}})}{1 - v} dv = \frac{1}{16} [ψ (\frac{5}{8}) - ψ (\frac{1}{8})]$
	Answered by Ar Brandon last updated on 30/Mar/21
	$I=∫_0 ^1 (x/(1+x^8 ))dx=(1/2)∫_0 ^1 ((2x)/(1+x^8 ))dx I=(1/2)∫_0 ^1 (dt/(t^4 +1))=(1/4)∫_0 ^1 (((t^2 +1)−(t^2 −1))/(t^4 +1))dt =(1/4)∫_0 ^1 {((t^2 +1)/(t^4 +1))−((t^2 −1)/(t^4 +1))}dt=(1/4)∫_0 ^1 {((1+(1/t^2 ))/(t^2 +(1/t^2 )))−((1−(1/t^2 ))/(t^2 +(1/t^2 )))}dt =(1/4){∫_0 ^1 ((1+(1/t^2 ))/((t−(1/t))^2 +2))dt−∫_0 ^1 ((1−(1/t^2 ))/((t+(1/t))^2 −2))dt} =(1/4){∫_(−∞) ^0 (du/(u^2 +2))−∫_(+∞) ^2 (dv/(v^2 −2))} =(1/4){[(1/( (√2)))arctan((u/( (√2))))]_(−∞) ^0 +[(1/(2(√2)))ln∣(((√2)+v)/( (√2)−v))∣]_(+∞) ^2 } =(1/4){(0+(π/(2(√2))))+(1/( 2(√2)))(ln∣(((√2)+2)/( (√2)−2))∣)}=(π/(8(√2)))+(1/( 8(√2)))ln∣((1+(√2))/( 1−(√2)))∣$
	$I = \int_{0}^{1} \frac{x}{1 + x^{8}} dx = \frac{1}{2} \int_{0}^{1} \frac{2 x}{1 + x^{8}} dx$ $I = \frac{1}{2} \int_{0}^{1} \frac{dt}{t^{4} + 1} = \frac{1}{4} \int_{0}^{1} \frac{(t^{2} + 1) - (t^{2} - 1)}{t^{4} + 1} dt$ $= \frac{1}{4} \int_{0}^{1} {\frac{t^{2} + 1}{t^{4} + 1} - \frac{t^{2} - 1}{t^{4} + 1}} dt = \frac{1}{4} \int_{0}^{1} {\frac{1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} - \frac{1 - \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}}} dt$ $= \frac{1}{4} {\int_{0}^{1} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} dt - \int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 2} dt}$ $= \frac{1}{4} {\int_{- \infty}^{0} \frac{du}{u^{2} + 2} - \int_{+ \infty}^{2} \frac{dv}{v^{2} - 2}}$ $= \frac{1}{4} {{[\frac{1}{\sqrt{2}} \arctan (\frac{u}{\sqrt{2}})]}_{- \infty}^{0} + {[\frac{1}{2 \sqrt{2}} \ln ∣ \frac{\sqrt{2} + v}{\sqrt{2} - v} ∣]}_{+ \infty}^{2}}$ $= \frac{1}{4} {(0 + \frac{π}{2 \sqrt{2}}) + \frac{1}{2 \sqrt{2}} (\ln ∣ \frac{\sqrt{2} + 2}{\sqrt{2} - 2} ∣)} = \frac{π}{8 \sqrt{2}} + \frac{1}{8 \sqrt{2}} \ln ∣ \frac{1 + \sqrt{2}}{1 - \sqrt{2}} ∣$
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