let F(x) =∫_x ^(1/x) ((arctan(t))/t)dt 1) calculate (dF/dx)(x) 2) find F(x).


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Question Number 36438 by prof Abdo imad last updated on 02/Jun/18

$l e t F (x) = \int_{x}^{\frac{1}{x}} \frac{a r c t a n (t)}{t} d t$ $1) c a l c u l a t e \frac{d F}{d x} (x)$ $2) f i n d F (x) .$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

Commented by prof Abdo imad last updated on 03/Jun/18
$changement t =(1/u) give 2)F(x) = ∫_(1/x) ^x (((π/2) −arctan(u))/(1/u)) (−(du/u^2 )) = ∫_x ^(1/x) (((π/2) −arctan(u))/u)du = (π/2) ∫_x ^(1/x) (du/u) − ∫_x ^(1/x) ((arctan(u))/u) du ⇒ 2F(x) = (π/2)[ln∣u∣]_x ^(1/x) =(π/2){ −2ln(x)} = −π ln(x) ⇒ F(x) =−(π/2)ln(x) 1) (dF/dx)(x) = −(π/(2x)) with x>0$
$c h a n g e m e n t t = \frac{1}{u} g i v e$ $2) F (x) = \int_{\frac{1}{x}}^{x} \frac{\frac{π}{2} - a r c t a n (u)}{\frac{1}{u}} (- \frac{d u}{u^{2}})$ $= \int_{x}^{\frac{1}{x}} \frac{\frac{π}{2} - a r c t a n (u)}{u} d u$ $= \frac{π}{2} \int_{x}^{\frac{1}{x}} \frac{d u}{u} - \int_{x}^{\frac{1}{x}} \frac{a r c t a n (u)}{u} d u \Rightarrow$ $2 F (x) = \frac{π}{2} {[l n ∣ u ∣]}_{x}^{\frac{1}{x}} = \frac{π}{2} {- 2 l n (x)}$ $= - π l n (x) \Rightarrow F (x) = - \frac{π}{2} l n (x)$ $1) \frac{d F}{d x} (x) = - \frac{π}{2 x} w i t h x > 0$
Commented by prof Abdo imad last updated on 03/Jun/18

$a d d x > 0 t o Q .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
	$F′(x)=∫_x ^(1/x) (∂/∂x)(((tan^(−1) t)/t))dt+((tan^(−1) ((1/x)))/(1/x))((d((1/x)))/dx)− ((tan^(−1) (x))/x)(dx/dx) =0+((tan^(−1) ((1/x)))/(1/x))×((−1)/x^2 )−((tan^(−1) (x))/x) =((tan^(−1) ((1/x)))/(−x))−((tan^(−1) (x))/x) =((−1)/x){((tan^(−1) ((1/x))+tan^(−1) (x))/1)} =((−1)/x)tan^(−1) {((x+(1/x)))/(1−x.(1/x)))} =(1/x)×((Π/2)] (dF/dx)=(Π/2)(1/x) dF=(Π/2)(dx/x) F=(Π/2)lnx$
	$F^{'} (x) = \int_{x}^{\frac{1}{x}} \frac{\partial}{\partial x} (\frac{{t a n}^{- 1} t}{t}) d t + \frac{{t a n}^{- 1} (\frac{1}{x})}{\frac{1}{x}} \frac{d (\frac{1}{x})}{d x} -$ $\frac{{t a n}^{- 1} (x)}{x} \frac{d x}{d x}$ $= 0 + \frac{{t a n}^{- 1} (\frac{1}{x})}{\frac{1}{x}} \times \frac{- 1}{x^{2}} - \frac{{t a n}^{- 1} (x)}{x}$ $= \frac{{t a n}^{- 1} (\frac{1}{x})}{- x} - \frac{{t a n}^{- 1} (x)}{x}$ $= \frac{- 1}{x} {\frac{{t a n}^{- 1} (\frac{1}{x}) + {t a n}^{- 1} (x)}{1}}$ $= \frac{- 1}{x} {t a n}^{- 1} {\frac{x + \frac{1}{x})}{1 - x . \frac{1}{x}}}$ $= \frac{1}{x} \times (\frac{Π}{2}]$ $\frac{d F}{d x} = \frac{Π}{2} \frac{1}{x}$ $d F = \frac{Π}{2} \frac{d x}{x}$ $F = \frac{Π}{2} l n x$
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