calculate ∫_1 ^(+∞) (dx/(x(√(4+x^2 ))))


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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
$another way we use the changement x =2sh(t) ⇒ A =∫_(argsh((1/2))) ^(+∞) ((2ch(t)dt)/(2sh(t)2 ch(t))) =(1/2) ∫_(ln((1/2)+(√(1+(1/4))))) ^(+∞) (dt/((e^t −e^(−t) )/2)) =∫_(ln((1/2)+((√5)/2))) ^(+∞) (dt/(e^t −e^(−t) )) =_(e^t =u) ∫_((1+(√5))/2) ^(+∞) (du/(u(u−(1/u)))) =∫_((1+(√5))/2) ^(+∞) (du/(u^2 −1)) =(1/2) ∫_((1+(√5))/2) ^(+∞) {(1/(u−1)) −(1/(u+1))}du =(1/2)[ln∣((u−1)/(u+1))∣]_((1+(√5))/2) ^(+∞) =(1/2){ −ln∣((((1+(√5))/2)−1)/(((1+(√5))/2)+1))∣} =−(1/2)ln∣(((√5)−1)/((√5)+3))∣ =(1/2)ln((((√5)+3)/((√5)−1))) .$
$a n o t h e r w a y w e u s e t h e c h a n g e m e n t x = 2 s h (t) \Rightarrow$ $A = \int_{a r g s h (\frac{1}{2})}^{+ \infty} \frac{2 c h (t) d t}{2 s h (t) 2 c h (t)} = \frac{1}{2} \int_{l n (\frac{1}{2} + \sqrt{1 + \frac{1}{4})}}^{+ \infty} \frac{d t}{\frac{e^{t} - e^{- t}}{2}}$ $= \int_{l n (\frac{1}{2} + \frac{\sqrt{5}}{2})}^{+ \infty} \frac{d t}{e^{t} - e^{- t}} =_{e^{t} = u} \int_{\frac{1 + \sqrt{5}}{2}}^{+ \infty} \frac{d u}{u (u - \frac{1}{u})}$ $= \int_{\frac{1 + \sqrt{5}}{2}}^{+ \infty} \frac{d u}{u^{2} - 1} = \frac{1}{2} \int_{\frac{1 + \sqrt{5}}{2}}^{+ \infty} {\frac{1}{u - 1} - \frac{1}{u + 1}} d u$ $= \frac{1}{2} {[l n ∣ \frac{u - 1}{u + 1} ∣]}_{\frac{1 + \sqrt{5}}{2}}^{+ \infty} = \frac{1}{2} {- l n ∣ \frac{\frac{1 + \sqrt{5}}{2} - 1}{\frac{1 + \sqrt{5}}{2} + 1} ∣}$ $= - \frac{1}{2} l n ∣ \frac{\sqrt{5} - 1}{\sqrt{5} + 3} ∣ = \frac{1}{2} l n (\frac{\sqrt{5} + 3}{\sqrt{5} - 1}) .$
	Answered by Tanmay chaudhury last updated on 17/Jul/19

	$\int \frac{x d x}{x^{2} \sqrt{4 + x^{2}}}$ $t^{2} = 4 + x^{2} 2 t d t = 2 x d x$ $\int \frac{t d t}{(t^{2} - 4) t}$ $\frac{1}{4} \int \frac{(t + 2) - (t - 2)}{(t + 2) (t - 2)} d t$ $\frac{1}{4} [\int \frac{d t}{t - 2} - \int \frac{d t}{t + 2}]$ $\frac{1}{4} l n (\frac{t - 2}{t + 2}) + c$ $\frac{1}{4} l n (\frac{\sqrt{4 + x^{2}} - 2}{\sqrt{4 + x^{2}} + 2}) + c$ $\frac{1}{4} ∣ l n (\frac{\sqrt{4 + x^{2}} - 2}{\sqrt{4 + x^{2}} + 2}) ∣_{1}^{+ \infty}$ $\frac{1}{4} ∣ l n (\frac{1 - \frac{2}{\sqrt{4 + x^{2}} [}}{1 + \frac{2}{\sqrt{4 + x^{2}}}}) ∣_{1}^{+ \infty}$ $= \frac{1}{4} (l n (\frac{1 - 0}{1 + 0}) - l n (\frac{\sqrt{5} - 2}{\sqrt{5} + 2}))$ $= - \frac{1}{4} \times l n (\frac{\sqrt{5} - 2}{\sqrt{5} + 2})$
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