find-f-t-0-1-ln-1-tx-2-dx-with-t-gt-0-

Question Number 30774 by abdo imad last updated on 25/Feb/18

f i n d f (t) = \int_{0}^{1} l n (1 + {t x}^{2}) d x w i t h t > 0

Commented by abdo imad last updated on 26/Feb/18

l e t p u t \sqrt{t} x = u \Rightarrow f (t) = \int_{0}^{\sqrt{t}} l n (1 + u^{2}) \frac{d u}{\sqrt{t}}

= \frac{1}{\sqrt{t}} \int_{0}^{\sqrt{t}} l n (1 + u^{2}) d u b u t w e h a v e b y p a r t s

\int_{0}^{\sqrt{t}} l n (1 + u^{2}) d u = {[u l n (1 + u^{2})]}_{0}^{\sqrt{t}} - \int_{0}^{\sqrt{t}} u \frac{2 u}{1 + u^{2}} d u

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

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

= \sqrt{t} l n (1 + t) - 2 \sqrt{t} + 2 a r c t a n (\sqrt{t}) \Rightarrow

f (t) = l n (1 + t) + \frac{2 a r c t a n (\sqrt{t})}{\sqrt{t}} - 2 .

Answered by sma3l2996 last updated on 25/Feb/18

l e t u = t x

f (t) = \int_{0}^{t} l n (1 + u^{2}) d u

= {[u l n (1 + u^{2})]}_{0}^{t} - 2 \int_{0}^{t} \frac{u^{2}}{1 + u^{2}} d u

= t l n (1 + t^{2}) - 2 \int_{0}^{t} (1 - \frac{1}{1 + u^{2}}) d u

= t l n (1 + t^{2}) - 2 {[u - {t a n}^{- 1} (u)]}_{0}^{t}

f (t) = t l n (1 + t^{2}) - 2 t + 2 {t a n}^{- 1} (t)

Commented by abdo imad last updated on 26/Feb/18

i f u = t x w e h a v e l n (1 + {t x}^{2}) \neq l n (1 + u^{2}) a n d y o u r a n s w e r

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