calculate-1-dx-x-2-4-x-2-

Question Number 36754 by prof Abdo imad last updated on 05/Jun/18

c a l c u l a t e \int_{1}^{+ \infty} \frac{d x}{x^{2} \sqrt{4 + x^{2}}} .

Commented by abdo mathsup 649 cc last updated on 05/Jun/18

c h a n g e m e n t x = 2 s h (t) g i v e

I = \int_{a r g s h (\frac{1}{2})}^{+ \infty} \frac{1}{4 {s h}^{2} t . 2 c h t} \frac{2 c h (t) d t}{}

= \frac{1}{4} \int_{a r g s h (\frac{1}{2})}^{+ \infty} \frac{d t}{\frac{c h (2 t) - 1}{2}}

= \frac{1}{2} \int_{a r g s h (\frac{1}{2})}^{+ \infty} \frac{d t}{\frac{e^{2 t} + e^{- 2 t}}{2} - 1}

= \int_{a r g s h (\frac{1}{2})}^{+ \infty} \frac{d t}{e^{2 t} + e^{- 2 t} - 2} = \int_{l n (\frac{1}{2} + \frac{\sqrt{5}}{2})}^{+ \infty} \frac{d t}{e^{2 t} + e^{- 2 t} - 2}

=_{e^{2 t} = u} \int_{{(\frac{1}{2} + \frac{\sqrt{5}}{2})}^{2}}^{+ \infty} \frac{1}{u + \frac{1}{u} - 2} \frac{d u}{2 u}

= \frac{1}{2} \int_{\frac{6 + 2 \sqrt{5}}{4}}^{+ \infty} \frac{d u}{u^{2} + 1} = \frac{1}{2} {[a r c t a n u]}_{\frac{6 + 2 \sqrt{5}}{4}}^{+ \infty}

= \frac{π}{4} - \frac{1}{2} a r c t a n (\frac{6 + 2 \sqrt{5}}{4}) .

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

l e t x = \frac{1}{t} s o d x = \frac{- 1}{t^{2}} d t

\int_{1}^{0} \frac{- d t}{t^{2} \times \frac{1}{t^{2}} \times \sqrt{4_{} + \frac{1}{t^{2}}}}

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

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

k^{2} = t^{2} + \frac{1}{4} 2 k d k = 2 t d t

= \frac{- 1}{2} \times \int_{\frac{\sqrt{5}}{2}}^{\frac{1}{2}} \frac{k d k}{k}

= \frac{- 1}{2} \times (\frac{1}{2} - \frac{\sqrt{5}}{2})

= \frac{\sqrt{5} - 1}{4}