let-f-x-2-x-x-1-find-f-x-f-1-x-dx-and-ln-f-x-f-1-x-dx-

Question Number 91620 by abdomathmax last updated on 02/May/20

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

f i n d \int \frac{f (x)}{f^{- 1} (x)} d x a n d \int l n (\frac{f (x)}{f^{- 1} (x)}) d x

Commented by mathmax by abdo last updated on 02/May/20

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

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

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

Δ = {(4 y + 1)}^{2} - 16 (y^{2} + 1) = 16 y^{2} + 8 y + 1 - 16 y^{2} - 16

= 8 y - 15 i f y ⩾ \frac{15}{8} x_{1} = \frac{4 y + 1 + \sqrt{8 y - 15}}{8} a n d x_{2} = \frac{4 y + 1 - \sqrt{8 y - 15}}{8}

w e m u s t h a v e x ⩾ 1 s o

x_{1} - 1 = \frac{4 y + 1 + \sqrt{8 y - 15}}{8} - 1 = \frac{4 y + 1 + \sqrt{8 y - 15} - 8}{8}

= \frac{4 y - 7 + \sqrt{8 y - 15}}{8} y ⩾ \frac{15}{8} \Rightarrow 4 y - 7 ⩾ \frac{15}{2} - 7 > 0 \Rightarrow x_{1} > 1 \Rightarrow

f^{- 1} (x) = \frac{4 x + 1 + \sqrt{8 x - 15}}{8} \Rightarrow

\int \frac{f (x)}{f^{- 1} (x)} d x = 8 \int \frac{2 x - \sqrt{x - 1}}{4 x + 1 + \sqrt{8 x - 15}} d x c h a n g e m e n t \sqrt{x - 1} = t g i v e

x - 1 = t^{2} \Rightarrow \int \frac{f (x)}{f^{- 1} (x)} d x = 8 \int \frac{2 (1 + t^{2}) - t}{4 (1 + t^{2}) + 1 + \sqrt{8 (1 + t^{2}) - 15}} (2 t) d t

= 16 \int \frac{2 t^{3} - t^{2} + 2 t}{5 + 4 t^{2} + \sqrt{8 t^{2} - 7}} d t c h a n g e m e n t t = \sqrt{\frac{7}{8}} c h (u) g i v e

I = 16 \int \frac{2 {(\sqrt{\frac{7}{8}})}^{3} {c h}^{3} (u) - \frac{7}{8} {c h}^{2} (u) + 2 \sqrt{\frac{7}{8}} c h (u)}{5 + \frac{7}{2} {c h}^{2} u + \sqrt{8} \times \sqrt{\frac{7}{8}} s h u} \times \sqrt{\frac{7}{8}} s h u d u

\dots b e c o n t i n u e d \dots