ind the value of f(a) =∫_0 ^(+∞) (dx/(x^2 +(√(a^2 +x^2 )))) dx witha>0 2)calculate f^′ (a) .


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Question Number 37902 by math khazana by abdo last updated on 19/Jun/18

$i n d t h e v a l u e o f f (a) = \int_{0}^{+ \infty} \frac{d x}{x^{2} + \sqrt{a^{2} + x^{2}}} d x$ $w i t h a > 0$ $2) c a l c u l a t e f^{^{'}} (a) .$
Commented by prof Abdo imad last updated on 19/Jun/18
$changement x=a tan t give f(a) = ∫_0 ^(π/2) (1/(a^2 tan^2 t + a cost)) a(1+tan^2 t)dt = ∫_0 ^(π/2) ((1+tan^2 t)/(a tant +cost)) dt chang. tan((t/2))=u give f(a) = ∫_0 ^1 ((1+( ((2u)/(1−u^2 )))^2 )/(a ((2u)/(1−u^2 )) +((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) = 2 ∫_0 ^1 (((1−u^2 )^2 +4u^2 )/((1+u^2 )(1−u^2 )^2 { ((2au)/(1−u^2 )) +((1−u^2 )/(1+u^2 ))}))du = 2 ∫_0 ^1 (((1−u^2 )^2 +4u^2 )/(2au(1−u^4 ) +(1−u^2 )^3 )) du =2 ∫_0 ^1 (((1+u^2 )^2 )/(2au −2au^5 −(u^(6 ) +3u^4 −3u^2 −1)))du =2∫_0 ^1 ((u^4 +2u^(2 ) +1)/(−u^6 −2au^5 +3u^2 +2au +1))du =−2 ∫_0 ^1 ((u^4 +2u^2 +1)/(u^6 +2au^5 −3u^2 −2au −1))du...be continued...$
$c h a n g e m e n t x = a t a n t g i v e$ $f (a) = \int_{0}^{\frac{π}{2}} \frac{1}{a^{2} {t a n}^{2} t + a c o s t} a (1 + {t a n}^{2} t) d t$ $= \int_{0}^{\frac{π}{2}} \frac{1 + {t a n}^{2} t}{a t a n t + c o s t} d t c h a n g . t a n (\frac{t}{2}) = u$ $g i v e$ $f (a) = \int_{0}^{1} \frac{1 + {(\frac{2 u}{1 - u^{2}})}^{2}}{a \frac{2 u}{1 - u^{2}} + \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}$ $= 2 \int_{0}^{1} \frac{{(1 - u^{2})}^{2} + 4 u^{2}}{(1 + u^{2}) {(1 - u^{2})}^{2} {\frac{2 a u}{1 - u^{2}} + \frac{1 - u^{2}}{1 + u^{2}}}} d u$ $= 2 \int_{0}^{1} \frac{{(1 - u^{2})}^{2} + 4 u^{2}}{2 a u (1 - u^{4}) + {(1 - u^{2})}^{3}} d u$ $= 2 \int_{0}^{1} \frac{{(1 + u^{2})}^{2}}{2 a u - 2 {a u}^{5} - (u^{6} + 3 u^{4} - 3 u^{2} - 1)} d u$ $= 2 \int_{0}^{1} \frac{u^{4} + 2 u^{2} + 1}{- u^{6} - 2 {a u}^{5} + 3 u^{2} + 2 a u + 1} d u$ $= - 2 \int_{0}^{1} \frac{u^{4} + 2 u^{2} + 1}{u^{6} + 2 {a u}^{5} - 3 u^{2} - 2 a u - 1} d u . . . b e$ $c o n t i n u e d . . .$
	Answered by ajfour last updated on 19/Jun/18

	$l e t x = a \tan θ$ $f (a) = \int_{0}^{π / 2} \frac{\sec^{2} θ d θ}{a \tan^{2} θ + \sec θ}$ $= \int_{0}^{π / 2} \frac{d θ}{a \sin^{2} θ + \cos θ}$ $l e t \tan \frac{θ}{2} = t$ $= \int_{0}^{1} \frac{\frac{2 d t}{1 + t^{2}}}{a {(\frac{2 t}{1 + t^{2}})}^{2} + \frac{1 - t^{2}}{1 + t^{2}}}$ $= \int_{0}^{1} \frac{2 (1 + t^{2}) d t}{4 {a t}^{2} + 1 - t^{4}}$ $l e t t^{4} - 4 {a t}^{2} + 1 = 0$ $\Rightarrow t^{2} = \frac{4 a \pm \sqrt{16 a^{2} - 4}}{2}$ $s a y α, β = 2 a \pm \sqrt{4 a^{2} - 1}$ $f (a) = - \int_{0}^{1} \frac{1 + t^{2}}{(t^{2} - α) (t^{2} - β)} d t$ $= - \frac{α + 1}{α - β} \int_{0}^{1} \frac{d t}{t^{2} - α} + \frac{1 + β}{α - β} \int_{0}^{1} \frac{d t}{t^{2} - β}$ $= - \frac{α + 1}{α - β} \times \frac{1}{2 \sqrt{α}} \ln ∣ \frac{t - \sqrt{α}}{t + \sqrt{α}} ∣_{0}^{1}$ $+ \frac{1 + β}{α - β} \times \frac{1}{2 \sqrt{β}} \ln ∣ \frac{t - \sqrt{β}}{t + \sqrt{β}} ∣_{0}^{1}$ $n o t s a t i s f a c t o r y, i g u e s s!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18

	$\int_{0}^{1} \frac{2 (\frac{1}{t^{2}} + 1) d t}{4 a + \frac{1}{t^{2}} - t^{2}}$ $\int_{0}^{1} \frac{2 d (t - \frac{1}{t})}{4 a - (t - \frac{1}{t}) (t + \frac{1}{t})}$ $2 \int_{0}^{1} \frac{d (t - \frac{1}{t})}{4 a - (t - \frac{1}{t}) \sqrt{{(t - \frac{1}{t})}^{2} + 4}}$ $2 \int_{- \infty}^{0} \frac{d k}{4 a - k \sqrt{k^{2} + 4}}$ $c o n t d$
	Answered by MJS last updated on 19/Jun/18

	$this is the method :$ $\int \frac{d x}{x^{2} + \sqrt{x^{2} + a^{2}}} = \int \frac{x^{2}}{x^{4} - x^{2} - a^{2}} d x - \int \frac{\sqrt{x^{2} + a^{2}}}{x^{4} - x^{2} - a^{2}} d x$ $\int \frac{x^{2}}{x^{4} - x^{2} - a^{2}} d x =$ $= \underset{i = 1}{\sum^{4}} \int \frac{N_{i}}{x - r_{i}} d x = \underset{i = 1}{\sum^{4}} N_{i} \ln ∣ x - r_{i} ∣$ $\int \frac{\sqrt{x^{2} + a^{2}}}{x^{4} - x^{2} - a^{2}} d x =$ $[t = \frac{x}{\sqrt{x^{2} + a^{2}}} \to d x = \frac{d t}{a^{2}} \sqrt{{(x^{2} + a^{2})}^{3}}]$ $= \frac{1}{a^{2}} \int \frac{d t}{t^{4} + \frac{1}{a^{2}} t^{2} - \frac{1}{a^{2}}} =$ $= \frac{N_{5}}{a^{2}} (\int \frac{d t}{t - r_{5}} - \int \frac{d t}{t + r_{5}}) + \frac{N_{6}}{a^{2}} \int \frac{d t}{t^{2} + r_{6}} =$ $[r_{6} > 0]$ $= \frac{N_{5}}{a^{2}} \ln ∣ \frac{t - r_{5}}{t + r_{5}} ∣ + \frac{N_{6}}{a^{2} \sqrt{r_{6}}} \arctan \frac{t}{\sqrt{r_{6}}} =$ $= \frac{N_{5}}{a^{2}} \ln ∣ \frac{\frac{x}{\sqrt{x^{2} + a^{2}}} - r_{5}}{\frac{x}{\sqrt{x^{2} + a^{2}}} + r_{5}} ∣ + \frac{N_{6}}{a^{2} \sqrt{r_{6}}} \arctan \frac{x}{\sqrt{r_{6} (x^{2} + a^{2})}}$
	Commented by MJS last updated on 20/Jun/18

	$r_{1} = - \frac{\sqrt{2}}{2} \sqrt{1 - \sqrt{4 a^{2} + 1}}$ $r_{2} = - \frac{\sqrt{2}}{2} \sqrt{1 + \sqrt{4 a^{2} + 1}}$ $r_{3} = \frac{\sqrt{2}}{2} \sqrt{1 - \sqrt{4 a^{2} + 1}}$ $r_{4} = \frac{\sqrt{2}}{2} \sqrt{1 + \sqrt{4 a^{2} + 1}}$ $r_{5} = \frac{\sqrt{2}}{2 a} \sqrt{- 1 + \sqrt{4 a^{2} + 1}}$ $r_{6} = \frac{1 + \sqrt{4 a^{2} + 1}}{2 a^{2}}$ $N_{1} = \frac{\sqrt{2 - 2 \sqrt{4 a^{2} + 1}}}{4 \sqrt{4 a^{2} + 1}}$ $N_{2} = - \frac{\sqrt{2 + 2 \sqrt{4 a^{2} + 1}}}{4 \sqrt{4 a^{2} + 1}}$ $N_{3} = - \frac{\sqrt{2 - 2 \sqrt{4 a^{2} + 1}}}{4 \sqrt{4 a^{2} + 1}}$ $N_{4} = \frac{\sqrt{2 + 2 \sqrt{4 a^{2} + 1}}}{4 \sqrt{4 a^{2} + 1}}$ $N_{5} = \frac{a^{3} \sqrt{2}}{2 \sqrt{(4 a^{2} + 1) (- 1 + \sqrt{4 a^{2} + 1})}}$ $N_{6} = - \frac{a^{2}}{\sqrt{4 a^{2} + 1}}$
	Commented by math khazana by abdo last updated on 20/Jun/18

	$t h a n k y o u s i r M j s$
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