fint-f-t-0-1-ln-1-tx-2-x-2-dx-

Question Number 55280 by Abdo msup. last updated on 20/Feb/19

f i n t f (t) = \int_{0}^{1} \frac{l n (1 + {t x}^{2})}{x^{2}} d x .

Commented by maxmathsup by imad last updated on 20/Feb/19

w e h a v e f^{^{'}} (t) = \int_{0}^{1} \frac{\partial}{\partial t} (\frac{l n (1 + {t x}^{2})}{x^{2}}) d x = \int_{0}^{1} \frac{1}{x^{2}} \frac{x^{2}}{1 + {t x}^{2}} d x

= \int_{0}^{1} \frac{d x}{1 + {t x}^{2}} s o i f t ⩾ 0 w e g e t f^{^{'}} (t) =_{\sqrt{t} x = u} \int_{0}^{\sqrt{t}} \frac{d u}{\sqrt{t} (1 + u^{2})}

= \frac{1}{\sqrt{t}} {[a r c t a n (u)]}_{0}^{\sqrt{t}} = \frac{a r c t a n (\sqrt{t})}{\sqrt{t}} \Rightarrow f (t) = \int_{0}^{t} \frac{a r c t a n (\sqrt{u})}{\sqrt{u}} d u + λ

λ = f (0) = 0 \Rightarrow f (t) = \int_{0}^{t} \frac{a r c t a n (\sqrt{u})}{\sqrt{u}} d u =_{\sqrt{u} = x} \int_{0}^{\sqrt{t}} \frac{a r c t a n x}{x} (2 x) d x

= 2 \int_{0}^{\sqrt{t}} a r c t a n x d x b y p a r t s f (t) = 2 {{[x a r c t a n x]}_{0}^{\sqrt{t}} - \int_{0}^{\sqrt{t}} \frac{x}{1 + x^{2}} d x}

= 2 {\sqrt{t} a r c t a n (\sqrt{t}) - {[\frac{1}{2} l n (1 + x^{2})]}_{0}^{\sqrt{t}} \Rightarrow f (t) = 2 \sqrt{t} a r c t a n (\sqrt{t}) - l n (1 + t) .

i t t < 0 w e g e t f^{^{'}} (t) = \int_{0}^{1} \frac{d x}{1 - (- t) x^{2}} = \int_{0}^{1} \frac{d x}{(1 - \sqrt{- t} x) (1 + \sqrt{- t} x)}

= \frac{1}{2} \int_{0}^{1} (\frac{1}{1 - \sqrt{- t} x} + \frac{1}{1 + \sqrt{- t} x}) d x = \frac{1}{2 \sqrt{- t}} {\int_{0}^{1} \frac{\sqrt{- t}}{1 - \sqrt{- t} x} + \int_{0}^{1} \frac{\sqrt{- t}}{1 + \sqrt{- t} x} d x}

= \frac{1}{2 \sqrt{- t}} {[l n ∣ 1 + \sqrt{- t} x ∣ - l n ∣ 1 - \sqrt{- t} x ∣]}_{x = 0}^{x = 1}

= \frac{1}{2 \sqrt{- t}} l n ∣ \frac{1 + \sqrt{- t}}{1 - \sqrt{- t}} ∣ \Rightarrow f (t) = \int \frac{1}{2 \sqrt{- t}} l n ∣ \frac{1 + \sqrt{- t}}{1 - \sqrt{- t}} ∣ d t + c

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

Answered by tm888 last updated on 20/Feb/19

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