0-dx-1-x-2-2x-x-2-1-pi-ln-3-2-3-solution-1-x-t-2-0-dt-1-t-4-2-0-1-dt-1-t-

Question Number 173149 by mnjuly1970 last updated on 07/Jul/22

\int_{0}^{\infty} \frac{d x}{\sqrt{1 + x} . (2 + 2 x + x^{2})} = \frac{1}{σ} (π - \ln (3 + 2 \sqrt{3}))

σ = ?

- - solution - -

Ω \overset{\sqrt{1 + x} = t}{=} 2 \int_{0}^{\infty} \frac{d t}{1 + t^{4}} = 2 \int_{0}^{1} \frac{d t}{1 + t^{4}}

Ψ = \int_{0}^{1} \frac{d t}{1 + t^{4}} = \frac{1}{2} \int_{0}^{1} \frac{1 + t^{2} - (t^{- 2} - 1)}{1 + t^{4}} d t

= \frac{1}{2} \int_{0}^{1} \frac{1 + t^{2}}{1 + t^{4}} d t + \frac{1}{2} \int_{0}^{1} \frac{1 - t^{2}}{1 + t^{4}} d t

Φ = \int_{0}^{1} \frac{1 + t^{2}}{1 + t^{4}} d t = \int_{0}^{1} \frac{t^{- 2} + 1}{t^{- 2} + t^{2}} d t

= \int_{0}^{1} \frac{1 + t^{- 2}}{{(t^{} - t^{- 1})}^{2} + 2} \overset{s u b}{=} {[\frac{1}{\sqrt{2}} {t a n}^{- 1} (t - t^{- 1})]}_{0}^{1} = \frac{π}{2 \sqrt{2}} *

ϕ = \int_{0}^{1} \frac{1 - t^{2}}{1 + t^{4}} d t = \int_{0}^{1} \frac{t^{- 2} - 1}{{(t + t^{- 1})}^{2} - 2} d t

\overset{t + \frac{1}{t} = u}{=} - \int_{2}^{\infty} \frac{d u}{(u - \sqrt{2}) (u + \sqrt{2})}

= \frac{- 1}{2 \sqrt{2}} {[\ln (\frac{u - \sqrt{2}}{u + \sqrt{2}})]}_{2}^{\infty} = \frac{1}{2 \sqrt{2}} \ln (\frac{2 - \sqrt{2}}{2 + \sqrt{2}})

ϕ = - \frac{1}{2 \sqrt{2}} \ln (3 + 2 \sqrt{2}) * *

(*) & (* *) :: Ω = 2 Ψ = (Φ + ϕ) = \frac{1}{2 \sqrt{2}} (π - \ln (3 + 2 \sqrt{2}) \dots m . n

Commented by Tawa11 last updated on 11/Jul/22

Great sir

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