f-x-3-x-1-x-1-f-x-f-1-x-

Question Number 41586 by MJS last updated on 09/Aug/18

f (x) = \sqrt{- 3 + \sqrt{\frac{x + 1}{x - 1}}}

\int f (x) = ?

\int f^{- 1} (x) = ?

Commented by prof Abdo imad last updated on 10/Aug/18

f (x) = y \Leftrightarrow x = f^{- 1} (y) \Rightarrow \sqrt{- 3 + \sqrt{\frac{x + 1}{x - 1}}} = y \Rightarrow

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

x + 1 = x {(y^{2} + 3)}^{2} - {(y^{2} + 3)}^{2} \Rightarrow

(1 - {(y^{2} + 3)}^{2}) x = - 1 - {(y^{2} + 3)}^{2} \Rightarrow

x = \frac{{(y^{2} + 3)}^{2} + 1}{{(y^{2} + 3)}^{2} - 1} = 1 + \frac{2}{{(y^{2} + 3)}^{2} - 1} \Rightarrow

f^{- 1} (x) = 1 + \frac{2}{{(x^{2} + 3)}^{2} - 1} \Rightarrow

\int f^{- 1} (x) d x = x + 2 \int \frac{d x}{{(x^{2} + 3)}^{2} - 1} + c b u t

\int \frac{2 d x}{{(x^{2} + 3)}^{2} - 1} = \int \frac{2 d x}{(x^{2} + 4) (x^{2} + 2)}

= \int {\frac{1}{x^{2} + 2} - \frac{1}{x^{2} + 4}} d x = \int \frac{d x}{x^{2} + 2} - \int \frac{d x}{x^{2} + 4}

\int \frac{d x}{x^{2} + 2} =_{x = \sqrt{2} t} \int \frac{\sqrt{2} d t}{2 (1 + t^{2})} = \frac{1}{\sqrt{2}} a r c t a n (\frac{x}{\sqrt{2}})

\int \frac{d x}{x^{2} + 4} =_{x = 2 t} \int \frac{2 d t}{4 (1 + t^{2})} = \frac{1}{2} a r c t a n (\frac{t}{2}) \Rightarrow

\int f^{- 1} (x) d x = x + \frac{1}{\sqrt{2}} a r c t a n (\frac{x}{\sqrt{2}}) + \frac{1}{2} a r c t a n (\frac{x}{2}) + c

Answered by MJS last updated on 09/Aug/18

\int \sqrt{- 3 + \sqrt{\frac{x + 1}{x - 1}}} d x =

[t = \sqrt{- 3 + \sqrt{\frac{x + 1}{x - 1}}} \to d x = - \frac{8 t (t^{2} + 3)}{{(t^{2} + 2)}^{2} {(t^{2} + 4)}^{2}} d t]

= - 8 \int \frac{t^{2} (t^{2} + 3)}{{(t^{2} + 2)}^{2} {(t^{2} + 4)}^{2}} d t =

= - 8 \int (- \frac{1}{2 {(t^{2} + 2)}^{2}} + \frac{1}{{(t^{2} + 4)}^{2}} + \frac{1}{4 (t^{2} + 2)} - \frac{1}{4 (t^{2} + 4)}) d t =

= 4 \int \frac{d t}{{(t^{2} + 2)}^{2}} - 8 \int \frac{d t}{{(t^{2} + 4)}^{2}} - 2 \int \frac{d t}{t^{2} + 2} + 2 \int \frac{d t}{t^{2} + 4} =

[\int \frac{d x}{{(x^{2} + p)}^{2}} = \frac{x}{2 p (x^{2} + p)} + \frac{\sqrt{p^{3}}}{2 p^{3}} \arctan \frac{x \sqrt{p}}{p}]

[\int \frac{d x}{x^{2} + p} = \frac{\sqrt{p}}{p} \arctan \frac{x \sqrt{p}}{p}]

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

= \frac{2 t}{(t^{2} + 2) (t^{2} + 4)} + \frac{1}{2} (\arctan \frac{t}{2} - \sqrt{2} \arctan \frac{t \sqrt{2}}{2}) =

= \sqrt{(\sqrt{x + 1} - 3 \sqrt{x - 1}) \sqrt{{(x - 1)}^{3}}} + \frac{1}{2} (\arctan (\frac{1}{2} \sqrt{- 3 + \sqrt{\frac{x + 1}{x - 1}}}) - \sqrt{2} \arctan (\frac{1}{2} \sqrt{- 6 + 2 \sqrt{\frac{x + 1}{x - 1}}})) + C

Answered by ajfour last updated on 10/Aug/18

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

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

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

I = \int [1 + \frac{2}{{(x^{2} + 3)}^{2} - 1}] d x

= x + 2 I_{1}

I_{1} = \int \frac{d x}{{(x^{2} + 3)}^{2} - 1} = \int \frac{d x}{(x^{2} + 2) (x^{2} + 4)}

= \frac{1}{2} \int \frac{d x}{x^{2} + 2} - \frac{1}{2} \int \frac{d x}{x^{2} + 4}

h e n c e

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

- \frac{1}{2} \tan^{- 1} \frac{x}{2} + c .

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