0-dx-x-1-x-2-2-let-x-tan-t-dx-sec-2-t-dt-0-pi-2-sec-2-t-dt-tan-t-sec-t-2-0-pi-2-dt-sin-t-1-2-0-pi-2-dt-cos-1-2-t-sin-1-2-t-4-0-

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 \dots 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

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 \dots

Commented by mathmax by abdo last updated on 24/Mar/20

a a h g o o d s i r \dots

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 ?