calculate-0-1-x-x-1-dx-

Question Number 37893 by abdo mathsup 649 cc last updated on 19/Jun/18

c a l c u l a t e \int_{0}^{1} \sqrt{x + \sqrt{x + 1}} d x .

Commented by math khazana by abdo last updated on 21/Jun/18

l e t \sqrt{x + \sqrt{x + 1}} = t \Rightarrow x + \sqrt{x + 1} = t^{2} \Rightarrow

\sqrt{x + 1} = t^{2} - x \Rightarrow x + 1 = {(t^{2} - x)}^{2} \Rightarrow

x = t^{4} - 2 {x t}^{2} + x^{2} - 1 \Rightarrow x^{2} - (2 t^{2} + 1) x + t^{4} - 1 = 0 \Rightarrow

Δ = {(2 t^{2} + 1)}^{2} - 4 (t^{4} - 1)

= 4 t^{4} + 4 t^{2} + 1 - 4 t^{4} + 4 = 4 t^{2} + 5 \Rightarrow

x_{1} = \frac{2 t^{2} + 1 + \sqrt{4 t^{2} + 5}}{2}

x_{2} = \frac{2 t^{2} + 1 - \sqrt{4 t^{2} + 5}}{2} w i t h c o n d i t i o n t ⩾ 0 a n d x ⩾ - 1

a n d t^{2} - x ⩾ 0

t^{2} - x_{2} = t^{2} - \frac{2 t^{2} + 1 - \sqrt{4 t^{2} + 5}}{2}

= \frac{- 1 + \sqrt{4 t^{2} + 5}}{2} ⩾ 0

x_{2} + 1 = \frac{2 t^{2} + 1 - \sqrt{4 t^{2} + 5}}{2} + 1

= \frac{2 t^{2} + 3 - \sqrt{4 t^{2} + 5}}{2} ⩾ 0 s o w e t a k e

x = \frac{2 t^{2} + 1 - \sqrt{4 t^{2} + 5}}{2} \Rightarrow \frac{d x}{d t} = 2 t - \frac{1}{2} \frac{8 t}{2 \sqrt{4 t^{2} + 5}}

= 2 t - \frac{2 t}{\sqrt{4 t^{2} + 5}} \Rightarrow

\int_{0}^{1} \sqrt{x + \sqrt{x + 1}} d x = \int_{1}^{\sqrt{1 + \sqrt{2}}} t (2 t - \frac{2 t}{\sqrt{4 t^{2} + 5}}) d t

= 2 \int_{1}^{\sqrt{1 + \sqrt{2}}} t^{2} d t - 2 \int_{1}^{\sqrt{1 + \sqrt{2}}} \frac{t^{2}}{\sqrt{4 t^{2} + 5}} d t

= 2 {[\frac{t^{3}}{3}]}_{1}^{\sqrt{1 + \sqrt{2}}} - \frac{1}{2} \int_{1}^{\sqrt{1 + \sqrt{2}}} \frac{4 t^{2} + 5 - 5}{\sqrt{4 t^{2} + 5}} d t

= 2 ((1 + \sqrt{2}) \sqrt{1 + \sqrt{2}} - \frac{1}{3}) - \frac{1}{2} \int_{1}^{\sqrt{1 + \sqrt{2}}} \sqrt{4 t^{2} + 5} d t

+ \frac{5}{2} \int_{1}^{\sqrt{1 + \sqrt{2}}} \frac{d t}{\sqrt{4 t^{2} + 5}} c h a n g e m e n t

2 t = \sqrt{5} s h (x) g i v e

\int_{1}^{\sqrt{1 + \sqrt{2}}} \sqrt{4 t^{2} + 5} d t = \int_{a r g s h (\frac{2}{\sqrt{5}})}^{a r s h (\frac{2}{\sqrt{5}} \sqrt{1 + \sqrt{2}})} \sqrt{5} c h (x) \frac{\sqrt{5}}{2} c h x d x

= \frac{5}{2} \int_{α}^{β} \frac{1 + c h (2 x)}{2} d x = \frac{5}{4} (β - α) + \frac{5}{8} {[s h (2 x)]}_{α}^{β}

= \frac{5}{4} (β - α) + \frac{5}{8} (s h (2 β) - s h (2 α)) \dots .

Answered by MJS last updated on 19/Jun/18

\int \sqrt{x + \sqrt{x + 1}} d x =

[t = \sqrt{x + 1} \to d x = 2 \sqrt{x + 1} d t]

= 2 \int t \sqrt{t^{2} + t - 1} d t =

= \int (2 t + 1) \sqrt{t^{2} + t - 1} d t - \int \sqrt{t^{2} + t - 1} d t =

\int (2 t + 1) \sqrt{t^{2} + t - 1} d t =

[u = t^{2} + t - 1 \to d t = \frac{d u}{2 t + 1}]

= \int \sqrt{u} d u = \frac{2}{3} \sqrt{u^{3}} = \frac{2}{3} \sqrt{{(t^{2} + t - 1)}^{3}} =

= \frac{2}{3} \sqrt{{(x + \sqrt{x + 1})}^{3}}

\int \sqrt{t^{2} + t - 1} d t = \int \sqrt{{(t + \frac{1}{2})}^{2} - \frac{5}{4}} d t =

= \frac{1}{2} \int \sqrt{{(2 t + 1)}^{2} - 5} d t =

[v = 2 t + 1 \to d t = \frac{d v}{2}]

= \frac{1}{4} \int \sqrt{v^{2} - 5} d v =

= \frac{1}{4} (\frac{1}{2} (v \sqrt{v^{2} - 5} - 5 \ln (v + \sqrt{v^{2} - 5}))) =

= \frac{1}{8} (2 t + 1) \sqrt{{(2 t + 1)}^{2} - 5} -

- \frac{5}{8} \ln (2 t + 1 + \sqrt{{(2 t + 1)}^{2} - 5}) =

= \frac{1}{4} (1 + 2 \sqrt{x + 1}) \sqrt{x + \sqrt{x + 1}} -

- \frac{5}{8} \ln (1 + 2 \sqrt{x + 1} + 2 \sqrt{x + \sqrt{x + 1}})

= \frac{1}{12} \sqrt{x + \sqrt{x + 1}} (8 x - 3 + 2 \sqrt{x + 1}) + \frac{5}{8} \ln (1 + 2 \sqrt{x + 1} + 2 \sqrt{x + \sqrt{x + 1}}) + C

Commented by prof Abdo imad last updated on 19/Jun/18

t h a n k s s i r M j s y o u a r e r e a l l y t a l e n t e d .