0-1-1-x-4-1-x-6-dx-p-2-q-12-pi-p-q-

Question Number 172973 by mnjuly1970 last updated on 04/Jul/22

\int_{0}^{\infty} \frac{1}{(1 + x^{4}) (1 + x^{6})} d x = \frac{p \sqrt{2} - q}{12} π

p, q = ?

Answered by floor(10²Eta[1]) last updated on 04/Jul/22

x^{6} + 1 = {(x^{2})}^{3} + 1 = (x^{2} + 1) (x^{4} - x^{2} + 1)

I = \int_{0}^{\infty} \frac{dx}{(x^{4} + 1) (x^{2} + 1) (x^{4} - x^{2} + 1)}

= \int_{0}^{\infty} (\frac{x^{2} + 1}{2 (x^{4} + 1)} + \frac{1}{6 (x^{2} + 1)} - \frac{{2 x}^{2} - 1}{3 (x^{4} - x^{2} + 1)}) dx

= \frac{1}{2} \int_{0}^{\infty} \frac{x^{2} + 1}{x^{4} + 1} dx + \frac{1}{6} \int_{0}^{\infty} \frac{dx}{x^{2} + 1} - \frac{1}{3} \int_{0}^{\infty} \frac{{2 x}^{2} - 1}{x^{4} - x^{2} + 1} dx

{= arctg (x)]}_{0}^{\infty} = \frac{π}{2}

= \int_{0}^{\infty} \frac{x^{2} + 1}{{(x^{2} + 1)}^{2} - {2 x}^{2}} dx = \int_{0}^{\infty} \frac{x^{2} + 1}{(x^{2} + \sqrt{2} x + 1) (x^{2} - \sqrt{2} x + 1)} dx

= \frac{1}{2} \int_{0}^{\infty} (\frac{1}{x^{2} + \sqrt{2} x + 1} + \frac{1}{x^{2} - \sqrt{2} x + 1}) dx

= \frac{1}{2} \int_{0}^{\infty} \frac{dx}{{(x + \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}} + \frac{1}{2} \int_{0}^{\infty} \frac{dx}{{(x - \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}}

= \frac{\sqrt{2}}{2} {[arctg (x \sqrt{2} + 1)]}_{0}^{\infty} + \frac{\sqrt{2}}{2} {[arctg (x \sqrt{2} - 1)]}_{0}^{\infty}

= \frac{\sqrt{2} π}{2}

= x^{4} - x^{2} + 1 = {(x^{2} + 1)}^{2} - {3 x}^{2}

\int_{0}^{\infty} \frac{{2 x}^{2} - 1}{(x^{2} + \sqrt{3} x + 1) (x^{2} - \sqrt{3} x + 1)} dx

= \frac{1}{2 \sqrt{3}} \int_{0}^{\infty} (\frac{- 3 x - \sqrt{3}}{x^{2} + \sqrt{3} x + 1} + \frac{3 x - \sqrt{3}}{x^{2} - \sqrt{3} x + 1}) dx

= - \frac{\sqrt{3}}{2} \int_{0}^{\infty} \frac{x}{x^{2} + \sqrt{3} x + 1} dx - \frac{1}{2} \int_{0}^{\infty} \frac{dx}{x^{2} + \sqrt{3} x + 1} + \frac{\sqrt{3}}{2} \int_{0}^{\infty} \frac{x}{x^{2} - \sqrt{3} x + 1} dx - \frac{1}{2} \int_{0}^{\infty} \frac{dx}{x^{2} - \sqrt{3} x + 1}

= - \frac{\sqrt{3}}{2} \int_{0}^{\infty} \frac{x}{{(x + \frac{\sqrt{3}}{2})}^{2} + \frac{1}{4}} dx - \frac{1}{2} \int_{0}^{\infty} \frac{dx}{{(x + \frac{\sqrt{3}}{2})}^{2} + \frac{1}{4}} + \frac{\sqrt{3}}{2} \int_{0}^{\infty} \frac{x}{{(x - \frac{\sqrt{3}}{2})}^{2} + \frac{1}{4}} dx - \frac{1}{2} \int_{0}^{\infty} \frac{dx}{{(x - \frac{\sqrt{3}}{2})}^{2} + \frac{1}{4}}

= - 2 \sqrt{3} \int_{0}^{\infty} \frac{x}{{(2 x + \sqrt{3})}^{2} + 1} dx - 2 \int_{0}^{\infty} \frac{dx}{{(2 x + \sqrt{3})}^{2} + 1} + 2 \sqrt{3} \int_{0}^{\infty} \frac{x}{{(2 x - \sqrt{3})}^{2} + 1} dx - 2 \int_{0}^{\infty} \frac{dx}{{(2 x - \sqrt{3})}^{2} + 1}

u = 2 x \pm \sqrt{3} \Rightarrow du = 2 dx, x = \frac{u \mp \sqrt{3}}{2}

= - \frac{\sqrt{3}}{2} \int_{\sqrt{3}}^{\infty} \frac{u - \sqrt{3}}{u^{2} + 1} du - \int_{\sqrt{3}}^{\infty} \frac{du}{u^{2} + 1} + \frac{\sqrt{3}}{2} \int_{- \sqrt{3}}^{\infty} \frac{u + \sqrt{3}}{u^{2} + 1} du - \int_{- \sqrt{3}}^{\infty} \frac{du}{u^{2} + 1}

- \frac{\sqrt{3}}{2} \int_{\sqrt{3}}^{\infty} \frac{u}{u^{2} + 1} du + \frac{1}{2} \int_{\sqrt{3}}^{\infty} \frac{du}{u^{2} + 1} + \frac{\sqrt{3}}{2} \int_{- \sqrt{3}}^{\infty} \frac{u}{u^{2} + 1} du + \frac{1}{2} \int_{- \sqrt{3}}^{\infty} \frac{du}{u^{2} + 1}

- \frac{\sqrt{3}}{2} \int_{\sqrt{3}}^{\infty} \frac{u}{u^{2} + 1} du + \frac{\sqrt{3}}{2} \int_{- \sqrt{3}}^{\infty} \frac{u}{u^{2} + 1} du + \frac{π}{2}

t = u^{2} + 1 \Rightarrow dt = 2 udu

\frac{- \sqrt{3}}{4} \int_{4}^{\infty} \frac{dt}{t} + \frac{\sqrt{3}}{4} \int_{4}^{\infty} \frac{dt}{t} + \frac{π}{2} = \frac{π}{2}

\Rightarrow I = \frac{1}{2} + \frac{1}{6} - \frac{1}{3}

= \frac{1}{2} (\frac{π \sqrt{2}}{2}) + \frac{1}{6} (\frac{π}{2}) - \frac{1}{3} (\frac{π}{2})

= \frac{π \sqrt{2}}{4} - \frac{π}{12} = \frac{3 \sqrt{2} - 1}{12} π \Rightarrow p = 3, q = 1

Commented by mnjuly1970 last updated on 04/Jul/22

bravo sir bravo . .

Commented by Tawa11 last updated on 06/Jul/22

Great sir

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