calculate-0-dx-1-2-x-2-2-

Question Number 89714 by abdomathmax last updated on 18/Apr/20

c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(1 + \sqrt{2 + x^{2}})}^{2}}

Commented by mathmax by abdo last updated on 19/Apr/20

p a r a m e t r i c m e t h o d l e t f (t) = \int_{0}^{\infty} \frac{d x}{t + \sqrt{2 + x^{2}}} w i t h t > 0

f^{^{'}} (t) = - \int_{0}^{\infty} \frac{d x}{{(t + \sqrt{2 + x^{2}})}^{2}} \Rightarrow \int_{0}^{\infty} \frac{d x}{{(t + \sqrt{2 + x^{2}})}^{2}} = - f^{^{'}} (t)

c h a n g e m e n t x = \sqrt{2} s h (u) g i v e

f (t) = \int_{0}^{\infty} \frac{\sqrt{2} c h (u) d u}{t + \sqrt{2} c h (u)} = \sqrt{2} \int_{0}^{\infty} \frac{\frac{e^{u} + e^{- u}}{2}}{t + \sqrt{2} \times \frac{e^{u} + e^{- u}}{2}} d u

= \sqrt{2} \int_{0}^{\infty} \frac{e^{u} + e^{- u}}{2 t + \sqrt{2} e^{u} + \sqrt{2} e^{- u}} d u =_{e^{u} = z} \sqrt{2} \int_{1}^{\infty} \frac{z + z^{- 1}}{2 t + \sqrt{2} z + \sqrt{2} z^{- 1}} \times \frac{d z}{z}

= \sqrt{2} \int_{1}^{+ \infty} \frac{z + z^{- 1}}{2 t z + \sqrt{2} z^{2} + \sqrt{2}} d z

= \sqrt{2} \int_{1}^{+ \infty} \frac{z^{2} + 1}{z (\sqrt{2} z^{2} + 2 t z + \sqrt{2})} d z d e c o p o s i t i o n o f g (z) = \frac{z^{2} + 1}{z (\sqrt{2} z^{2} + 2 t z + \sqrt{2})}

\sqrt{2} z^{2} + 2 t z + \sqrt{2} = 0 \to Δ^{^{'}} = t^{2} - 2 < 0 i f t < \sqrt{2}

z_{1} = \frac{- t + i \sqrt{2 - t^{2}}}{\sqrt{2}} a n d z_{2} = \frac{- t - i \sqrt{2 - t^{2}}}{\sqrt{2}}

g (z) = \frac{z^{2} + 1}{\sqrt{2} z (z - z_{1}) (z - z_{2})} = \frac{a}{z} + \frac{b}{z - z_{1}} + \frac{c}{z - z_{2}}

a = \frac{1}{\sqrt{2} z_{1} z_{2}} = \frac{1}{\sqrt{2}}, b = \frac{z_{1}^{2} + 1}{\sqrt{2} z_{1} \sqrt{2} i \sqrt{2 - t^{2}}} = \frac{1 + z_{1}^{2}}{2 {i z}_{1} \sqrt{2 - t^{2}}}

b = \frac{z_{2}^{2} + 1}{\sqrt{2} z_{2} (- \sqrt{2} i \sqrt{2 - t^{2}})} = \frac{1 + z_{2}^{2}}{- 2 {i z}_{2} \sqrt{2 - t^{2}}} \Rightarrow

\int_{1}^{+ \infty} g (z) d z = \int_{1}^{+ \infty} (\frac{a}{z} + \frac{b}{z - z_{1}} + \frac{c}{z - z_{2}}) d z

= {[a l n ∣ z ∣ + b l n ∣ z - z_{1} ∣ + c l n ∣ z - z_{2} ∣]}_{1}^{+ \infty} r e s t t o s i m p l i f y b a n d c

b e c o n t i n u e d \dots

Answered by MJS last updated on 18/Apr/20

\int \frac{d x}{{(1 + \sqrt{x^{2} + 2})}^{2}} =

[t = \frac{x + \sqrt{x^{2} + 2}}{\sqrt{2}} \to d x = \frac{\sqrt{2 (x^{2} + 2)}}{x + \sqrt{x^{2} + 2}} d t]

= \sqrt{2} \int \frac{t^{2} + 1}{{(t^{2} + \sqrt{2} t + 1)}^{2}} d t =

[Ostrogradski]

= \frac{\sqrt{2} t + 2}{t^{2} + \sqrt{2} t + 1} + 2 \sqrt{2} \int \frac{d t}{t^{2} + \sqrt{2} t + 1} =

= \frac{\sqrt{2} t + 2}{t^{2} + \sqrt{2} t + 1} + 4 \arctan (\sqrt{2} t + 1) =

= \frac{x^{2} + x + 1 - x \sqrt{x^{2} + 2}}{x^{2} + 1} + 4 \arctan (x + 1 + \sqrt{x^{2} + 2}) + C

\Rightarrow

\underset{0}{\int^{\infty}} \frac{d x}{{(1 + \sqrt{x^{2} + 2})}^{2}} = \frac{π}{2} - 1

Commented by turbo msup by abdo last updated on 18/Apr/20

t h a n k y o u s i r .