find-x-x-1-x-1-dx-

Question Number 137004 by Mathspace last updated on 28/Mar/21

f i n d \int \frac{\sqrt{x}}{\sqrt{x - 1} + \sqrt{x + 1}} d x

Answered by aleks041103 last updated on 28/Mar/21

\frac{\sqrt{x}}{\sqrt{x - 1} + \sqrt{x + 1}} =

= \frac{\sqrt{x}}{\sqrt{x - 1} + \sqrt{x + 1}} \frac{\sqrt{x + 1} - \sqrt{x - 1}}{\sqrt{x + 1} - \sqrt{x - 1}} =

= \sqrt{x} (\sqrt{x + 1} - \sqrt{x - 1})

I (a) = \int \sqrt{x (x + 2 a)} d x =

= \int \sqrt{((x + a) - a) ((x + a) + a)} d x =

= \int \sqrt{{(x + a)}^{2} - a^{2}} d x =

= a^{2} \int \sqrt{{(1 + x / a)}^{2} - 1} d (1 + \frac{x}{a})

c o s h (t) = 1 + \frac{x}{a}

s i n h (t) d t = d (1 + \frac{x}{a})

I (a) = a^{2} \int \sqrt{{c o s h}^{2} (t) - 1} s i n h (t) d t =

= a^{2} \int {s i n h}^{2} (t) d t

{s i n h}^{2} (t) = c o s h (2 t) - 1

I (a) = a^{2} (\frac{1}{2} s i n h (2 t) - t)

t = a r c c o s h (1 + x / a) = l n (1 + \frac{x}{a} + \sqrt{{(1 + x / a)}^{2} - 1}) =

= l n (1 + a^{- 1} x + a^{- 1} \sqrt{x (x + 2 a)})

\frac{1}{2} s i n h (2 t) = s i n h (t) c o s h (t) =

= c o s h (t) \sqrt{{c o s h}^{2} (t) - 1} =

= (1 + x / a) \sqrt{{(1 + x / a)}^{2} - 1} =

= a^{- 2} (x + a) \sqrt{x (x + 2 a)}

I (a) = (x + a) \sqrt{x (x + 2 a)} - a^{2} l n (1 + a^{- 1} x + a^{- 1} \sqrt{x (x + 2 a)})

\int \frac{\sqrt{x}}{\sqrt{x - 1} + \sqrt{x + 1}} d x = \int \sqrt{x} (\sqrt{x + 1} - \sqrt{x - 1}) d x =

= I (1 / 2) - I (- 1 / 2) =

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

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

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

Answered by mathmax by abdo last updated on 29/Mar/21

thankx sir .

Answered by EDWIN88 last updated on 29/Mar/21

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

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

E = - \frac{1}{2} \int \sqrt{x^{2} - x} dx + \frac{1}{2} \int \sqrt{x^{2} + x} dx

E = - \frac{1}{2} \int \sqrt{{(x - \frac{1}{2})}^{2} - \frac{1}{4}} dx + \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} - \frac{1}{4}} dx

E_{1} = - \frac{1}{2} \int \sqrt{{(x - \frac{1}{2})}^{2} - \frac{1}{4}} dx

let x - \frac{1}{2} = \frac{1}{2} \sec t dx = \frac{1}{2} \sec t \tan t dt

E_{1} = - \frac{1}{2} \int \frac{1}{2} \tan t . \frac{1}{2} \sec t \tan t dt

E_{1} = - \frac{1}{8} \int \sec t (\sec^{2} t - 1) dt

E_{1} = - \frac{1}{8} \int \sec^{3} t - \sec t dt

similary to E_{2} = \frac{1}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} - \frac{1}{4}} dx

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