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Question Number 62129 by maxmathsup by imad last updated on 15/Jun/19

l e t f (x) = l n (x + 1 - 2 \sqrt{x})

1) f i n d D_{f}

2) d e t e r m i n e f^{- 1}

3) c a l c u l a t e \int f (x) d x a n d \int f^{- 1} (x) d x

Commented by arcana last updated on 16/Jun/19

1)

x + 1 - 2 \sqrt{x} > 0 \Rightarrow {(\sqrt{x} - 1)}^{2} > 0

n e c e s s a r y \sqrt{x} - 1 > 0 \lor \sqrt{x} - 1 < 0

\sqrt{x} > 1 \lor \sqrt{x} < 1

x > 1 \lor x < 1, x ⩾ 0

\Rightarrow D_{f} = [0, 1) \cup (1, + \infty) = R^{+} - {1}

Commented by kaivan.ahmadi last updated on 16/Jun/19

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

x = \sqrt{1 + \sqrt{e^{y}}} \Rightarrow f^{- 1} (x) = \sqrt{1 + \sqrt{e^{x}}}

Commented by maxmathsup by imad last updated on 16/Jun/19

1) x \in D_{f} \Leftrightarrow x ⩾ 0 a n d x - 2 \sqrt{x} + 1 > 0 \Leftrightarrow x ⩾ 0 a n d {(\sqrt{x} - 1)}^{2} > 0 \Leftrightarrow x ⩾ 0 a n d x \neq 1 \Rightarrow

D_{f} = [0, 1 [\cup] 1, + \infty [

2) f (x) = y \Leftrightarrow x = f^{- 1} (y)

f (x) = y \Rightarrow l n (x + 1 - 2 \sqrt{x}) = y \Rightarrow x + 1 - 2 \sqrt{x} = e^{y} \Rightarrow {(\sqrt{x} - 1)}^{2} = e^{y} \Rightarrow

∣ \sqrt{x} - 1 ∣ = \sqrt{e^{y}} b y e^{y} > 0 \Rightarrow x > 1 \Rightarrow \sqrt{x} - 1 = \sqrt{e^{y}} \Rightarrow \sqrt{x} = 1 + e^{\frac{y}{2}} \Rightarrow

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

3) l e t I = \int f (x) d x \Rightarrow I = \int l n (x + 1 - 2 \sqrt{x}) d x b y p a r t s

I = x l n (x + 1 - 2 \sqrt{x}) - \int x \frac{1 - \frac{1}{\sqrt{x}}}{x + 1 - 2 \sqrt{x}} d x

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

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

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

= 2 \frac{t^{2}}{2} + 2 t + 2 l n ∣ t - 1 ∣ + c = t^{2} + 2 t + 2 l n ∣ t - 1 ∣ + c

= x + 2 \sqrt{x} + 2 l n ∣ \sqrt{x} - 1 ∣ + c \Rightarrow

I = x l n (x + 1 - 2 \sqrt{x}) - x - 2 \sqrt{x} - 2 l n ∣ \sqrt{x} - 1 ∣ + c .

Commented by maxmathsup by imad last updated on 16/Jun/19

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

= x + 4 e^{\frac{x}{2}} + e^{x} + c .