find f(t)=∫_0 ^1 ln(1+tx^2 )dx with t>0


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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 byabdo 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 byabdo imad last updated on 26/Feb/18

	$i f u = t x w e h a v e l n (1 + {t x}^{2}) \neq l n (1 + u^{2}) a n d y o u r a n s w e r$ $s i r s m a 3 i s n o t c o r e c t . . .$
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