find-0-1-dx-x-x-1-3-

Question Number 145183 by mathmax by abdo last updated on 03/Jul/21

find \int_{0}^{1} \frac{dx}{{(\sqrt{x} + \sqrt{x + 1})}^{3}}

Commented by justtry last updated on 03/Jul/21

Answered by mindispower last updated on 03/Jul/21

x = {s h}^{2} (t)

\Leftrightarrow \int \frac{s h (2 t) d t}{e^{3 t}}

= \frac{1}{2 e^{t}} - \frac{e^{- 5 t}}{2}

= - \frac{e^{- t}}{2} + \frac{e^{- 5 t}}{10} + c, t = a r g s h (\sqrt{x})

Answered by mathmax by abdo last updated on 03/Jul/21

Ψ = \int_{0}^{1} \frac{dx}{{(\sqrt{x} + \sqrt{x + 1})}^{3}} changement x = {sh}^{2} t give

sht = \sqrt{x} \Rightarrow t = argsh (\sqrt{x}) = \log (\sqrt{x} + \sqrt{1 + x}) \Rightarrow

Ψ = \int_{0}^{\log (1 + \sqrt{2})} \frac{2 sht cht}{{(sht + cht)}^{3}} dt = \int_{0}^{\log (1 + \sqrt{2})} \frac{sh (2 t)}{e^{3 t}} dt

= \int_{0}^{\log (1 + \sqrt{2})} e^{- 3 t} . \frac{e^{2 t} - e^{- 2 t}}{2} dt = \frac{1}{2} \int_{0}^{\log (1 + \sqrt{2})} (e^{- t} - e^{- 5 t}) dt

= \frac{1}{2} {[- e^{- t} + \frac{1}{5} e^{- 5 t}]}_{0}^{\log (1 + \sqrt{2})}

Ψ = \frac{1}{2} (- \frac{1}{1 + \sqrt{2}} + \frac{1}{5 {(1 + \sqrt{2})}^{5}} + 1 - \frac{1}{5})