∫ _0 ^∞ (dx/((x+(√(1+x^2 )))^2 )) let x = tan t ⇒dx=sec^2 t dt ∫_0 ^(π/2) ((sec^2 t dt)/((tan t+sec t)^2 )) = ∫_0 ^(π/2) (dt/((sin t+1)^2 )) = ∫_0 ^(π/2) (dt/((cos (1/2)t+sin (1/2)t)^4 )) = ∫_0 ^(π/2) (dt/(4cos^4 ((1/2)t−(π/4)))) = (1/4)∫_0 ^(π/2) sec^4 ((1/2)t−(π/4)) dt [ let (1/2)t−(π/4)= u] = (1/4)∫_(−(π/4)) ^0 sec^4 u ×2du =(1/2)∫ _(−(π/4)) ^0 (tan^2 u+1) d(tan u) = (1/2) [(1/3)tan^3 u + tan u ]_(−(π/4)) ^0 = (1/2) [ 0−(−(1/3)−1)]= (2/3)


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Question Number 85676 by john santu last updated on 24/Mar/20

$\int \underset{0}{\overset{\infty}{}} \frac{d x}{{(x + \sqrt{1 + x^{2}})}^{2}}$ $l e t x = \tan t \Rightarrow d x = \sec^{2} t d t$ $\underset{0}{\int^{\frac{π}{2}}} \frac{\sec^{2} t d t}{{(\tan t + \sec t)}^{2}} =$ $\underset{0}{\int^{\frac{π}{2}}} \frac{d t}{{(\sin t + 1)}^{2}} = \underset{0}{\int^{\frac{π}{2}}} \frac{d t}{{(\cos \frac{1}{2} t + \sin \frac{1}{2} t)}^{4}}$ $= \underset{0}{\int^{\frac{π}{2}}} \frac{d t}{{4 \cos}^{4} (\frac{1}{2} t - \frac{π}{4})}$ $= \frac{1}{4} \underset{0}{\int^{\frac{π}{2}}} \sec^{4} (\frac{1}{2} t - \frac{π}{4}) d t$ $[l e t \frac{1}{2} t - \frac{π}{4} = u]$ $= \frac{1}{4} \underset{- \frac{π}{4}}{\int^{0}} \sec^{4} u \times 2 d u$ $= \frac{1}{2} \int \underset{- \frac{π}{4}}{\overset{0}{}} (\tan^{2} u + 1) d (\tan u)$ $= \frac{1}{2} {[\frac{1}{3} \tan^{3} u + \tan u]}_{- \frac{π}{4}}^{0}$ $= \frac{1}{2} [0 - (- \frac{1}{3} - 1)] = \frac{2}{3}$
Commented by jagoll last updated on 24/Mar/20

$i try by Euler substitution$ $let \sqrt{1 + x^{2}} = x + t$ $1 + x^{2} = x^{2} + 2 xt + t^{2}$ $2 xt + t^{2} = 1 \Rightarrow x = \frac{1 - t^{2}}{2 t}$ $dx = \frac{- t^{2} - 1}{{2 t}^{2}} dt$ $\int \frac{1}{{(\frac{1 - t^{2}}{2 t} + \frac{1 + t^{2}}{2 t})}^{2}} \times (\frac{- t^{2} - 1}{{2 t}^{2}}) dt$ $\int \frac{t^{2}}{{2 t}^{2}} \times (- t^{2} - 1) dt = - \frac{1}{2} \int (t^{2} + 1) dt$ $- \frac{1}{2} [\frac{1}{3} t^{3} + t] = - \frac{1}{6} t [t^{2} + 3]$ $- \frac{1}{6} (\sqrt{1 + x^{2}} - x) ({2 x}^{2} + 4 - 2 x \sqrt{1 + x^{2}})$ $\lim_{x \to \infty} - \frac{1}{6} (\sqrt{1 + x^{2}} - x) ({2 x}^{2} + 4 - 2 x \sqrt{1 + x^{2}}) + \frac{4}{6}$ $= \frac{2}{3}$
Commented by john santu last updated on 24/Mar/20

$g o o d$
Commented by sakeefhasan05@gmail.com last updated on 24/Mar/20

$\int_{0}^{\infty} \frac{1}{{(x + \sqrt{1 + x^{2}})}^{n}} dx = \frac{n}{n^{2} - 1}$ $common solution$
Commented by sakeefhasan05@gmail.com last updated on 24/Mar/20

$comment pls$
Commented by john santu last updated on 24/Mar/20

$w a w . . . i t g e n e r a l l y s o l u t i o n s i r ?$
Commented by sakeefhasan05@gmail.com last updated on 24/Mar/20

$yeah . n = 3 check (\frac{3}{8}) pls try$
Commented by jagoll last updated on 24/Mar/20

$but n \neq 1 ?$
Commented by sakeefhasan05@gmail.com last updated on 24/Mar/20

$sry \frac{n}{(n^{2} - 1)}, [n^{2} - 1 \neq 0] so (n \neq 1) & (n \neq - 1)$
Commented by sakeefhasan05@gmail.com last updated on 24/Mar/20

$thank you for direct me$
Commented by mathmax by abdo last updated on 24/Mar/20
$let A_n =∫_0 ^∞ (dx/((x+(√(1+x^2 )))^n )) we do the changement x=sh(t) ⇒ A_n =∫_0 ^∞ ((ch(t))/((sh(t)+ch(t))^n ))dt =∫_0 ^∞ ((ch(t))/((((e^t −e^(−t) )/2)+((e^t +e^(−t) )/2))^n ))dt =∫_0 ^∞ e^(−nt) (((e^t +e^(−t) )/2))dt =(1/2) ∫_0 ^∞ ( e^((−n+1)t) +e^(−(n+1)t) )dt =(1/2)[(1/(1−n))e^((1−n)t) −(1/(n+1))e^(−(n+1)t) ]_0 ^(+∞) =(1/2){−(1/(1−n))+(1/(n+1))} =(1/2)((1/(n+1))+(1/(n−1))) =(1/2)(((2n)/(n^2 −1))) ⇒ A_n =(n/(n^2 −1)) (n>1) so ∫_0 ^∞ (dx/((x+(√(1+x^2 )))^2 )) =(2/(2^2 −1)) =(2/3)$
$l e t A_{n} = \int_{0}^{\infty} \frac{d x}{{(x + \sqrt{1 + x^{2}})}^{n}} w e d o t h e c h a n g e m e n t x = s h (t) \Rightarrow$ $A_{n} = \int_{0}^{\infty} \frac{c h (t)}{{(s h (t) + c h (t))}^{n}} d t = \int_{0}^{\infty} \frac{c h (t)}{{(\frac{e^{t} - e^{- t}}{2} + \frac{e^{t} + e^{- t}}{2})}^{n}} d t$ $= \int_{0}^{\infty} e^{- n t} (\frac{e^{t} + e^{- t}}{2}) d t = \frac{1}{2} \int_{0}^{\infty} (e^{(- n + 1) t} + e^{- (n + 1) t}) d t$ $= \frac{1}{2} {[\frac{1}{1 - n} e^{(1 - n) t} - \frac{1}{n + 1} e^{- (n + 1) t}]}_{0}^{+ \infty}$ $= \frac{1}{2} {- \frac{1}{1 - n} + \frac{1}{n + 1}} = \frac{1}{2} (\frac{1}{n + 1} + \frac{1}{n - 1}) = \frac{1}{2} (\frac{2 n}{n^{2} - 1}) \Rightarrow$ $A_{n} = \frac{n}{n^{2} - 1} (n > 1) s o$ $\int_{0}^{\infty} \frac{d x}{{(x + \sqrt{1 + x^{2}})}^{2}} = \frac{2}{2^{2} - 1} = \frac{2}{3}$
Commented by john santu last updated on 24/Mar/20

$t h a n k y o u s i r f o r y o u r s h o r t c u t$
Commented by john santu last updated on 24/Mar/20

$i n d o n e s i a n s i r$
Commented by mathmax by abdo last updated on 24/Mar/20

$w h e r e a r e y o u f r o m s i r j o h n . . .$
Commented by mathmax by abdo last updated on 24/Mar/20

$a a h g o o d s i r . . .$
Commented by john santu last updated on 25/Mar/20

$w h y s i r ? i f y o u w h e r e y o u$ $c o m e s i r ?$
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