let f(x) =((1−(√(x−1)))/(2+(√(x−1)))) 1) find f^(−1) (x) 2) find ∫ f(x)dx 3) find ∫ f^(−1) (x)dx 4) find ∫ (dx/(f^(−1) (x))) .


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Question Number 43809 by maxmathsup by imad last updated on 15/Sep/18

$l e t f (x) = \frac{1 - \sqrt{x - 1}}{2 + \sqrt{x - 1}}$ $1) f i n d f^{- 1} (x)$ $2) f i n d \int f (x) d x$ $3) f i n d \int f^{- 1} (x) d x$ $4) f i n d \int \frac{d x}{f^{- 1} (x)} .$
Commented by maxmathsup by imad last updated on 17/Sep/18

$1) f i s d e f i n e d o n [1, + \infty [l e t f (x) = y \Rightarrow x = f^{- 1} (y)$ $f (x) = y \Rightarrow \frac{1 - \sqrt{x - 1}}{2 + \sqrt{x - 1}} = y \Rightarrow 1 - \sqrt{x - 1} = 2 y + y \sqrt{x - 1} \Rightarrow 1 - 2 y = (1 + y) \sqrt{x - 1} \Rightarrow$ $\sqrt{x - 1} = \frac{1 - 2 y}{1 + y} \Rightarrow x - 1 = {(\frac{1 - 2 y}{1 + y})}^{2} \Rightarrow x = {(\frac{2 y - 1}{y + 1})}^{2} + 1 \Rightarrow f^{- 1} (x) = {(\frac{2 x - 1}{x + 1})}^{2} + 1$ $2) c h a n g e m e n t \sqrt{x - 1} = t g i v e x - 1 = t^{2} \Rightarrow$ $\int f (x) d x = \int \frac{1 - t}{2 + t} (2 t) d t = 2 \int \frac{t^{2} - t}{t + 2} d t = 2 \int \frac{t^{2} + 2 t - 3 t}{t + 2} d t$ $= 2 \int \frac{t (t + 2)}{t + 2} d t - 6 \int \frac{t + 2 - 2}{t + 2} d t = t^{2} - 6 t + 12 l n ∣ t + 2 ∣ + c \Rightarrow$ $\int f (x) d x = x - 1 - 6 \sqrt{x - 1} + 12 l n ∣ \sqrt{x - 1} + 2 ∣ + c .$
Commented by maxmathsup by imad last updated on 17/Sep/18

$3) \int f^{- 1} (x) d x = \int (\frac{4 x^{2} - 4 x + 1}{x^{2} + 2 x + 1} + 1) d x = \int \frac{4 x^{2} - 4 x + 1 + x^{2} + 2 x + 1}{x^{2} + 2 x + 1} d x$ $= \int \frac{5 x^{2} - 2 x + 2}{x^{2} + 2 x + 1} d x = \int \frac{5 (x^{2} + 2 x + 1) - 12 x - 3}{x^{2} + 2 x + 1} d x$ $= 5 x - \int \frac{12 x - 3}{x^{2} + 2 x + 1} d x b u t \int \frac{12 x - 3}{x^{2} + 2 x + 1} d x = \int \frac{12 x - 3}{{(x + 1)}^{2}} d x$ $= \int \frac{12 (x + 1) - 15}{{(x + 1)}^{2}} d x = 12 \int \frac{d x}{x + 1} - 15 \int \frac{d x}{{(x + 1)}^{2}} = 12 l n ∣ x + 1 ∣ + \frac{15}{x + 1} + c \Rightarrow$ $\int f^{- 1} (x) d x = 5 x - 12 l n ∣ x + 1 ∣ - \frac{15}{x + 1} + c .$
Commented by maxmathsup by imad last updated on 17/Sep/18

$4) \int \frac{d x}{f^{- 1} (x)} = \int \frac{x^{2} + 2 x + 1}{5 x^{2} - 2 x + 2} d x = \frac{1}{5} \int \frac{5 x^{2} + 10 x + 5}{5 x^{2} - 2 x + 2} d x$ $= \frac{1}{5} \int \frac{5 x^{2} - 2 x + 2 + 12 x + 3}{5 x^{2} - 2 x + 2} d x = \frac{1}{5} x + \frac{1}{5} \int \frac{12 x + 3}{5 x^{2} - 2 x + 2} d x b u t$ $\int \frac{12 x + 3}{5 x^{2} - 2 x + 2} d x = \int \frac{10 x - 2 + 2 x + 5}{5 x^{2} - 2 x + 2} d x$ $= l n ∣ 5 x^{2} - 2 x + 2 ∣ + \int \frac{2 x + 5}{5 x^{2} - 2 x + 2}$ $= l n ∣ 5 x^{2} - 2 x + 2 ∣ + \frac{1}{5} \int \frac{10 x - 2 + 27}{5 x^{2} - 2 x + 2} d x$ $= \frac{6}{5} l n ∣ 5 x^{2} - 2 x + 2 ∣ + \frac{27}{25} \int \frac{d x}{x^{2} - \frac{2}{5} x + \frac{2}{5}} b u t$ $\int \frac{d x}{x^{2} - \frac{2}{5} x + \frac{2}{5}} = \int \frac{d x}{x^{2} - \frac{2}{5} x + \frac{1}{25} + \frac{10}{25} - \frac{1}{25}} = \int \frac{d x}{{(x - \frac{1}{5})}^{2} + \frac{9}{25}}$ $=_{x - \frac{1}{5} = \frac{3}{5} u} \frac{25}{9} \int \frac{1}{1 + u^{2}} \frac{3}{5} d u = \frac{5}{3} a r c t a n (\frac{5 x - 1}{3}) + c \Rightarrow$ $\int \frac{d x}{f^{- 1} (x)} = \frac{x}{5} + \frac{6}{25} l n ∣ 5 x^{2} - 2 x + 2 ∣ + \frac{27}{125} \frac{5}{3} a r c t a n (\frac{5 x - 1}{3}) + c$ $= \frac{x}{5} + \frac{6}{25} l n ∣ 5 x^{2} - 2 x + 2 ∣ + \frac{9}{25} a r c t a n (\frac{5 x - 1}{3}) + c .$
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