calculate-3-dx-x-2-x-2-

Question Number 35686 by prof Abdo imad last updated on 22/May/18

c a l c u l a t e \int_{\sqrt{3}}^{+ \infty} \frac{d x}{x \sqrt{2 + x^{2}}} .

Commented by prof Abdo imad last updated on 22/May/18

l e t p u t I = \int_{\sqrt{3}}^{+ \infty} \frac{d x}{x \sqrt{2 + x^{2}}} c h a n g e m e n t

x = \sqrt{2} s h (t) g i v e t = a r g s h (\frac{x}{\sqrt{2}}) a n d

I = \int_{a r g s h (\frac{\sqrt{3}}{\sqrt{2}})}^{+ \infty} \frac{\sqrt{2} c h (t) d t}{\sqrt{2} s h (t) \sqrt{2} \sqrt{1 + {s h t}^{2}}}

= \int_{l n (\frac{\sqrt{3}}{\sqrt{2}} + \sqrt{1 + \frac{3}{2}})}^{+ \infty} \frac{d t}{\sqrt{2} s h (t)}

= \frac{1}{\sqrt{2}} \int_{l n (\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}})}^{+ \infty} 2 \frac{d t}{e^{t} - e^{- t}}

= \sqrt{2} \int_{l n (\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}})}^{+ \infty} \frac{d t}{e^{t} - e^{- t}}

=_{e^{t} = x} \sqrt{2} \int_{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}}}^{+ \infty} \frac{1}{x - \frac{1}{x}} \frac{d x}{x}

I = \sqrt{2} \int_{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}}}^{+ \infty} \frac{d x}{x^{2} - 1}

= \frac{\sqrt{2}}{2} \int_{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}}}^{+ \infty} {\frac{1}{x - 1} - \frac{1}{x + 1}} d x

= \frac{\sqrt{2}}{2} {[l n ∣ \frac{x - 1}{x + 1} ∣]}_{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}}}^{+ \infty} = - \frac{\sqrt{2}}{2} l n (\frac{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}} - 1}{\frac{\sqrt{3} + \sqrt{5}}{\sqrt{2}} + 1})

= - \frac{\sqrt{2}}{2} l n (\frac{\sqrt{3} + \sqrt{5} - \sqrt{2}}{\sqrt{3} + \sqrt{5} + \sqrt{2}})

Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18

= \int_{\sqrt{3}}^{\infty} \frac{x d x}{x^{2} \sqrt{2 + x^{2}}}

t^{2} = 2 + x^{2} 2 t d t = 2 x d x

= \int_{\sqrt{5}}^{\infty} \frac{t d t}{(t^{2} - 2) t}

= \int_{\sqrt{5}}^{\infty} \frac{d t}{(t^{2} - 2)}

= \frac{1}{2 \sqrt{2}} ∣ l n (\frac{t - \sqrt{2}}{t + \sqrt{2}}) ∣_{\sqrt{5}}^{\infty}

= \frac{1}{2 \sqrt{2}} ∣ l n (\frac{1 - \frac{\sqrt{2}}{t}}{1 + \frac{\sqrt{2}}{t}}) ∣_{\sqrt{5}}^{\infty}

= \frac{1}{2 \sqrt{2}} {0 - l n (\frac{1 - \frac{\sqrt{2}}{\sqrt{5}}}{1 + \frac{\sqrt{2}}{\sqrt{5}}})}

= \frac{1}{2 \sqrt{2}} {- l n (\frac{\sqrt{5} - \sqrt{2}}{\sqrt{5 + \sqrt{2}_{}}}}