calculate-0-1-tanh-1-x-1-x-2-dx-

Question Number 202866 by mnjuly1970 last updated on 04/Jan/24

c a l c u l a t e \dots

ϕ = \int_{0}^{1} \frac{{t a n h}^{- 1} (x)}{{(1 + x)}^{2}} d x = ?

Answered by Mathspace last updated on 06/Jan/24

x = t h (t) = \frac{s h t}{c h t} = \frac{e^{t} - e^{- t}}{e^{t} + e^{- t}}

\frac{d x}{d t} = \frac{{c h}^{2} t - {s h}^{2} t}{{c h}^{2} t} = \frac{1}{{c h}^{2} t}

Φ = \int_{0}^{a r g t h 1} \frac{t}{{(1 + \frac{e^{t} - e^{- t}}{e^{t} + e^{- t}})}^{2}} \frac{d t}{{(\frac{e^{t} + e^{- t}}{2})}^{2}}

= 4 \int_{0}^{a r g t h (1)} \frac{t}{4 e^{2 t}} d t

= \int_{0}^{a r g t h (1)} t e^{- 2 t} d t

= {[- \frac{1}{2} t e^{- 2 t}]}_{0}^{a r g t h (1)} + \frac{1}{2} \int_{0}^{a r g t h 1} e^{- 2 t} d t

w e h a v e x = \frac{e^{t} - e^{- t}}{e^{t} + e^{- t}}

= \frac{e^{2 t} - 1}{e^{2 t} + 1} \Rightarrow {x e}^{2 t} + x = e^{2 t} - 1 \Rightarrow

(x - 1) e^{2 t} = - x - 1 \Rightarrow

e^{2 t} = \frac{1 + x}{1 - x} \Rightarrow t = \frac{1}{2} l n (\frac{1 + x}{1 - x})

x \to 1^{-} \Rightarrow t \to + \infty \Rightarrow

Φ = {[- \frac{1}{2} {t e}^{- 2 t}]}_{0}^{\infty} - \frac{1}{4} {[e^{- 2 t}]}_{0}^{\infty}

= \frac{1}{4}