let f(x)=∫_0 ^1 ((ln(1+xt^2 ))/t^2 )dt with x ∈R 1) find a explicit form of f(x) 2)calculate ∫_0 ^1 ((ln(1+t^2 ))/t^2 )dt 3)calculate ∫_0 ^1 ((ln(1+2t^2 ))/t^2 )dt 4) calculate ∫


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Question Number 44201 by abdo.msup.com last updated on 23/Sep/18

$l e t f (x) = \int_{0}^{1} \frac{l n (1 + {x t}^{2})}{t^{2}} d t w i t h x \in R$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{1} \frac{l n (1 + t^{2})}{t^{2}} d t$ $3) c a l c u l a t e \int_{0}^{1} \frac{l n (1 + 2 t^{2})}{t^{2}} d t$ $4) c a l c u l a t e \int_{0}^{1} \frac{l n (1 - t^{2})}{t^{2}} d t$
Commented by maxmathsup by imad last updated on 25/Sep/18
$1) we have f^′ (x) = ∫_0 ^1 (t^2 /(t^2 (1+xt^2 )))dt = ∫_0 ^1 (dt/(1 +xt^2 )) case 1 x>0 changement t(√x)=u give f^′ (x)= ∫_0 ^(√x) (1/(1+u^2 )) (du/(√x)) =(1/(√x)) arctan((√x)) ⇒ f(x) =∫_0 ^x ((arctan((√t)))/(√t)) dt +c but c=f(0)=0 ⇒ f(x) =∫_0 ^x ((actan((√t)))/(√t))dt =_((√t)=u) ∫_0 ^(√x) ((arctan(u))/u) (2u)du =2 ∫_0 ^(√x) arctanu du by parts f(x) =2{ [u arctanu]_0 ^(√x) −∫_0 ^(√x) (u/(1+u^2 ))du} =2{(√x)arctan((√x)) −[(1/2)ln(1+u^2 )]_0 ^(√x) } =2{ (√x)arctan((√x)) −(1/2)ln(1+x)} ⇒f(x)=2(√x)arctan((√x))−ln(1+x) case 2 x<0 ⇒f^′ (x)= ∫_0 ^1 (dt/(1−(−x)t^2 )) =_(t(√(−x))=u) ∫_0 ^(√(−x)) (1/(1−u^2 )) (du/(√(−x))) =(1/(2(√(−x)))){ ∫_0 ^(√(−x)) { (1/(1−u)) +(1/(1+u))}du} = (1/(2(√(−x))))[ln∣((1+u)/(1−u))∣]_0 ^(√(−x)) ⇒ f^′ (x) = (1/(2(√(−x))))ln∣((1+(√(−x)))/(1−(√(−x))))∣ ⇒f(x) = ∫_x ^0 (1/(2(√(−t))))ln∣((1+(√(−t)))/(1−(√(−t))))∣ +c c=f(0) =0 ⇒f(x)= ∫_x ^0 (1/(2(√(−t))))ln∣((1+(√(−t)))/(1−(√(−t))))∣ dt changement (√(−t))=u give f(x) = ∫_(√(−x)) ^0 (1/(2u))ln∣((1+u)/(1−u))∣(−2u)du = ∫_0 ^(√(−x)) ln∣1+u∣du−∫_0 ^(√(−x)) ln∣1−u∣ du =∫_0 ^(√(−x)) ln(1+u)du −∫_0 ^(√(−x)) ln(1−u)du but ∫_0 ^(√(−x)) ln(1+u)du =_(1+u =α) ∫_1 ^(1+(√(−x))) ln(α)dα [αln(α)−α]_1 ^(1+(√(−x))) =(1+(√(−x)))ln(1+(√(−x)))−(1+(√(−x))) +1 =(1+(√(−x)))ln(1+(√(−x))) −(√(−x)).also ∫_0 ^(√(−x)) ln(1−u)du =_(1−u =α) −∫_1 ^(1−(√(−x))) ln(α)dα =−[αln(α) −α]_1 ^(1−(√(−x))) =−{(1−(√(−x)))ln(1−(√(−x)))−(1−(√(−x))) +1} =−(1−(√(−x)))ln(1−(√(−x))) −(√(−x)) ⇒ f(x)=(1+(√(−x)))ln(1+(√(−x))) +(1−(√(−x)))ln(1−(√(−x))).$
$1) w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{t^{2}}{t^{2} (1 + {x t}^{2})} d t = \int_{0}^{1} \frac{d t}{1 + {x t}^{2}}$ $c a s e 1 x > 0 c h a n g e m e n t t \sqrt{x} = u g i v e f^{^{'}} (x) = \int_{0}^{\sqrt{x}} \frac{1}{1 + u^{2}} \frac{d u}{\sqrt{x}}$ $= \frac{1}{\sqrt{x}} a r c t a n (\sqrt{x}) \Rightarrow f (x) = \int_{0}^{x} \frac{a r c t a n (\sqrt{t})}{\sqrt{t}} d t + c b u t c = f (0) = 0 \Rightarrow$ $f (x) = \int_{0}^{x} \frac{a c t a n (\sqrt{t})}{\sqrt{t}} d t =_{\sqrt{t} = u} \int_{0}^{\sqrt{x}} \frac{a r c t a n (u)}{u} (2 u) d u = 2 \int_{0}^{\sqrt{x}} a r c t a n u d u$ $b y p a r t s f (x) = 2 {{[u a r c t a n u]}_{0}^{\sqrt{x}} - \int_{0}^{\sqrt{x}} \frac{u}{1 + u^{2}} d u}$ $= 2 {\sqrt{x} a r c t a n (\sqrt{x}) - {[\frac{1}{2} l n (1 + u^{2})]}_{0}^{\sqrt{x}}}$ $= 2 {\sqrt{x} a r c t a n (\sqrt{x}) - \frac{1}{2} l n (1 + x)} \Rightarrow f (x) = 2 \sqrt{x} a r c t a n (\sqrt{x}) - l n (1 + x)$ $c a s e 2 x < 0 \Rightarrow f^{^{'}} (x) = \int_{0}^{1} \frac{d t}{1 - (- x) t^{2}} =_{t \sqrt{- x} = u} \int_{0}^{\sqrt{- x}} \frac{1}{1 - u^{2}} \frac{d u}{\sqrt{- x}}$ $= \frac{1}{2 \sqrt{- x}} {\int_{0}^{\sqrt{- x}} {\frac{1}{1 - u} + \frac{1}{1 + u}} d u} = \frac{1}{2 \sqrt{- x}} {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{\sqrt{- x}} \Rightarrow$ $f^{^{'}} (x) = \frac{1}{2 \sqrt{- x}} l n ∣ \frac{1 + \sqrt{- x}}{1 - \sqrt{- x}} ∣ \Rightarrow f (x) = \int_{x}^{0} \frac{1}{2 \sqrt{- t}} l n ∣ \frac{1 + \sqrt{- t}}{1 - \sqrt{- t}} ∣ + c$ $c = f (0) = 0 \Rightarrow f (x) = \int_{x}^{0} \frac{1}{2 \sqrt{- t}} l n ∣ \frac{1 + \sqrt{- t}}{1 - \sqrt{- t}} ∣ d t$ $c h a n g e m e n t \sqrt{- t} = u g i v e f (x) = \int_{\sqrt{- x}}^{0} \frac{1}{2 u} l n ∣ \frac{1 + u}{1 - u} ∣ (- 2 u) d u$ $= \int_{0}^{\sqrt{- x}} l n ∣ 1 + u ∣ d u - \int_{0}^{\sqrt{- x}} l n ∣ 1 - u ∣ d u = \int_{0}^{\sqrt{- x}} l n (1 + u) d u - \int_{0}^{\sqrt{- x}} l n (1 - u) d u$ $b u t \int_{0}^{\sqrt{- x}} l n (1 + u) d u =_{1 + u = α} \int_{1}^{1 + \sqrt{- x}} l n (α) d α$ ${[α l n (α) - α]}_{1}^{1 + \sqrt{- x}} = (1 + \sqrt{- x}) l n (1 + \sqrt{- x}) - (1 + \sqrt{- x}) + 1$ $= (1 + \sqrt{- x}) l n (1 + \sqrt{- x}) - \sqrt{- x} . a l s o$ $\int_{0}^{\sqrt{- x}} l n (1 - u) d u =_{1 - u = α} - \int_{1}^{1 - \sqrt{- x}} l n (α) d α$ $= - {[α l n (α) - α]}_{1}^{1 - \sqrt{- x}} = - {(1 - \sqrt{- x}) l n (1 - \sqrt{- x}) - (1 - \sqrt{- x}) + 1}$ $= - (1 - \sqrt{- x}) l n (1 - \sqrt{- x}) - \sqrt{- x} \Rightarrow$ $f (x) = (1 + \sqrt{- x}) l n (1 + \sqrt{- x}) + (1 - \sqrt{- x}) l n (1 - \sqrt{- x}) .$
Commented by maxmathsup by imad last updated on 25/Sep/18

$2) w e h a v e f (x) = \int_{0}^{1} \frac{l n (1 + {x t}^{2})}{t^{2}} d t \Rightarrow \int_{0}^{1} \frac{l n (1 + t^{2})}{t^{2}} d t$ $= f (1) = 2 a r c t a n (1) - l n (2) = \frac{2 π}{4} - l n (2) = \frac{π}{2} - l n (2) .$
Commented by maxmathsup by imad last updated on 25/Sep/18

$\int_{0}^{1} \frac{l n (1 + 2 t^{2})}{t^{2}} d t = f (2) = 2 \sqrt{2} a r c t a n (\sqrt{2}) - l n (3) .$
Commented by maxmathsup by imad last updated on 25/Sep/18

$4) \int_{0}^{1} \frac{l n (1 - t^{2})}{t^{2}} d t = f (- 1) = 2 l n (2) .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18
	$(df/dx)=∫_0 ^1 (∂/∂x)×{((ln(1+xt^2 ))/t^2 )}dt =∫_0 ^1 (1/t^2 )×((0+t^2 )/(1+xt^2 )) dt =∫_0 ^1 (dt/(x((1/x)+t^2 ))) =(1/x)×(1/(1/((√x) )))∣tan^(−1) ((t/(1/(√x_ ))))∣_0 ^1 =(1/((√x) ))tan^(−1) ((√x) ) df(x)=((tan^(−1) ((√x) ))/((√x) ))dx f(x)=∫((tan^(−1) ((√x) ))/((√x) ))dx y^2 =x dx=2ydy ∫((tan^(−1) (y) 2ydy)/y) =2∫tan^(−1) (y)dy =2ytan^(−1) (y)−∫(2/(1+y^2 ))×ydy =2ytan^(−1) (y)−ln(1+y^2 )+c =2(√x) tan^(−1) ((√x) )−ln(1+x)+c$
	$\frac{d f}{d x} = \int_{0}^{1} \frac{\partial}{\partial x} \times {\frac{l n (1 + {x t}^{2})}{t^{2}}} d t$ $= \int_{0}^{1} \frac{1}{t^{2}} \times \frac{0 + t^{2}}{1 + {x t}^{2}} d t$ $= \int_{0}^{1} \frac{d t}{x (\frac{1}{x} + t^{2})}$ $= \frac{1}{x} \times \frac{1}{\frac{1}{\sqrt{x}}} ∣ {t a n}^{- 1} (\frac{t}{\frac{1}{\sqrt{x_{}}}}) ∣_{0}^{1}$ $= \frac{1}{\sqrt{x}} {t a n}^{- 1} (\sqrt{x})$ $d f (x) = \frac{{t a n}^{- 1} (\sqrt{x})}{\sqrt{x}} d x$ $f (x) = \int \frac{{t a n}^{- 1} (\sqrt{x})}{\sqrt{x}} d x$ $y^{2} = x$ $d x = 2 y d y$ $\int \frac{{t a n}^{- 1} (y) 2 y d y}{y}$ $= 2 \int {t a n}^{- 1} (y) d y$ $= 2 {y t a n}^{- 1} (y) - \int \frac{2}{1 + y^{2}} \times y d y$ $= 2 {y t a n}^{- 1} (y) - l n (1 + y^{2}) + c$ $= 2 \sqrt{x} {t a n}^{- 1} (\sqrt{x}) - l n (1 + x) + c$
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