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

Question Number 64390 by mathmax by abdo last updated on 17/Jul/19

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

Commented by mathmax by abdo last updated on 18/Jul/19

l e t I = \int_{1}^{+ \infty} \frac{d x}{x \sqrt{4 + x^{2}}} c h a n g e m e n t x = 2 t a n θ g i v e

I = \int_{a r c t a n (\frac{1}{2})}^{\frac{π}{2}} \frac{2 (1 + {t a n}^{2} θ) d θ}{2 t a n θ \times 2 \sqrt{1 + {t a n}^{2} θ}}

= \frac{1}{2} \int_{a r c t a n (\frac{1}{2})}^{\frac{π}{2}} \frac{\sqrt{1 + {t a n}^{2} θ}}{t a n θ} d θ = \frac{1}{2} \int_{a r c t a n (\frac{1}{2})}^{\frac{π}{2}} \frac{1}{c o s θ \frac{s i n θ}{c o s θ}} d θ

= \frac{1}{2} \int_{a r c t a n (\frac{1}{2})}^{\frac{π}{2}} \frac{d θ}{s i n θ} =_{t a n (\frac{θ}{2}) = u} \frac{1}{2} \int_{t a n (\frac{1}{2} a r c t a n (\frac{1}{2}))}^{1} \frac{2 d u}{(1 + u^{2}) \frac{2 u}{1 + u^{2}}}

{= \int_{λ}^{1} \frac{d u}{2 u} = \frac{1}{2} l n ∣ u ∣]}_{λ}^{1} = - \frac{1}{2} l n λ = - \frac{1}{2} l n (t a n (\frac{1}{2} a r c t a n (\frac{1}{2}))

Commented by mathmax by abdo last updated on 18/Jul/19

a n o t h e r w a y w e u s e t h e c h a n g e m e n t x = 2 s h (t) \Rightarrow

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

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

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

= \frac{1}{2} {[l n ∣ \frac{u - 1}{u + 1} ∣]}_{\frac{1 + \sqrt{5}}{2}}^{+ \infty} = \frac{1}{2} {- l n ∣ \frac{\frac{1 + \sqrt{5}}{2} - 1}{\frac{1 + \sqrt{5}}{2} + 1} ∣}

= - \frac{1}{2} l n ∣ \frac{\sqrt{5} - 1}{\sqrt{5} + 3} ∣ = \frac{1}{2} l n (\frac{\sqrt{5} + 3}{\sqrt{5} - 1}) .

Answered by Tanmay chaudhury last updated on 17/Jul/19

\int \frac{x d x}{x^{2} \sqrt{4 + x^{2}}}

t^{2} = 4 + x^{2} 2 t d t = 2 x d x

\int \frac{t d t}{(t^{2} - 4) t}

\frac{1}{4} \int \frac{(t + 2) - (t - 2)}{(t + 2) (t - 2)} d t

\frac{1}{4} [\int \frac{d t}{t - 2} - \int \frac{d t}{t + 2}]

\frac{1}{4} l n (\frac{t - 2}{t + 2}) + c

\frac{1}{4} l n (\frac{\sqrt{4 + x^{2}} - 2}{\sqrt{4 + x^{2}} + 2}) + c

\frac{1}{4} ∣ l n (\frac{\sqrt{4 + x^{2}} - 2}{\sqrt{4 + x^{2}} + 2}) ∣_{1}^{+ \infty}

\frac{1}{4} ∣ l n (\frac{1 - \frac{2}{\sqrt{4 + x^{2}} [}}{1 + \frac{2}{\sqrt{4 + x^{2}}}}) ∣_{1}^{+ \infty}

= \frac{1}{4} (l n (\frac{1 - 0}{1 + 0}) - l n (\frac{\sqrt{5} - 2}{\sqrt{5} + 2}))

= - \frac{1}{4} \times l n (\frac{\sqrt{5} - 2}{\sqrt{5} + 2})