f(x) =∫_0 ^1 (dt/(x+e^t )) with 0≤x≤1 1) find aexplicit form of f(x) 2) calculate ∫_0 ^1 (dt/(2+e^t )) 3) find g(x) =∫_0 ^1 (dt/((x+e^t )^2 )) 4) calculate ∫_0 ^1 (dt/((1+e^t )^2 )) 5) give f^((n)) (x) at form of integrals 6) developp f at integr serie.


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Question Number 65290 by mathmax by abdo last updated on 27/Jul/19

$f (x) = \int_{0}^{1} \frac{d t}{x + e^{t}} w i t h 0 ⩽ x ⩽ 1$ $1) f i n d a e x p l i c i t f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{1} \frac{d t}{2 + e^{t}}$ $3) f i n d g (x) = \int_{0}^{1} \frac{d t}{{(x + e^{t})}^{2}}$ $4) c a l c u l a t e \int_{0}^{1} \frac{d t}{{(1 + e^{t})}^{2}}$ $5) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l s$ $6) d e v e l o p p f a t i n t e g r s e r i e .$
Commented by mathmax by abdo last updated on 28/Jul/19
$1) f(x) =∫_0 ^1 (dt/(x+e^t )) =∫_0 ^1 (e^(−t) /(1+xe^(−t) ))dt =−(1/x)∫_0 ^1 ((−xe^(−t) )/(1+xe^(−t) ))dt =−(1/x)[ln(1+xe^(−t) )]_(t=0) ^1 =−(1/x){ln(1+xe^(−1) )−ln(1+x)} ⇒ f(x) =((ln(1+x))/x) −((ln(1+xe^(−1) ))/x) if x≠0 2)∫_0 ^1 (dt/(2+e^t )) =f(2) =((ln(3))/2) −((ln(1+2e^(−1) ))/2) 3)we have f^′ (x) =−∫_0 ^1 (dt/((x+e^t )^2 )) =−g(x) ⇒g(x)=−f^′ (x) f^′ (x) =−(1/x^2 )ln(1+x)+(1/(x(1+x))) +(1/x^2 )ln(1+xe^(−1) )−(1/x)(e^(−1) /(1+xe^(−1) )) ⇒ g(x) =(1/x^2 )ln(1+x)−(1/(x(1+x)))−(1/x^2 )ln(1+xe^(−1) )+(e^(−1) /(x(1+xe^(−1) ))) 4)∫_0 ^1 (dt/((1+e^t )^2 )) =g(1) =ln(2)−(1/2) −ln(1+e^(−1) ) +(e^(−1) /(1+e^(−1) ))$
$1) f (x) = \int_{0}^{1} \frac{d t}{x + e^{t}} = \int_{0}^{1} \frac{e^{- t}}{1 + {x e}^{- t}} d t = - \frac{1}{x} \int_{0}^{1} \frac{- {x e}^{- t}}{1 + {x e}^{- t}} d t$ $= - \frac{1}{x} {[l n (1 + {x e}^{- t})]}_{t = 0}^{1} = - \frac{1}{x} {l n (1 + {x e}^{- 1}) - l n (1 + x)} \Rightarrow$ $f (x) = \frac{l n (1 + x)}{x} - \frac{l n (1 + {x e}^{- 1})}{x} i f x \neq 0$ $2) \int_{0}^{1} \frac{d t}{2 + e^{t}} = f (2) = \frac{l n (3)}{2} - \frac{l n (1 + 2 e^{- 1})}{2}$ $3) w e h a v e f^{^{'}} (x) = - \int_{0}^{1} \frac{d t}{{(x + e^{t})}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x)$ $f^{^{'}} (x) = - \frac{1}{x^{2}} l n (1 + x) + \frac{1}{x (1 + x)} + \frac{1}{x^{2}} l n (1 + {x e}^{- 1}) - \frac{1}{x} \frac{e^{- 1}}{1 + {x e}^{- 1}} \Rightarrow$ $g (x) = \frac{1}{x^{2}} l n (1 + x) - \frac{1}{x (1 + x)} - \frac{1}{x^{2}} l n (1 + {x e}^{- 1}) + \frac{e^{- 1}}{x (1 + {x e}^{- 1})}$ $4) \int_{0}^{1} \frac{d t}{{(1 + e^{t})}^{2}} = g (1) = l n (2) - \frac{1}{2} - l n (1 + e^{- 1}) + \frac{e^{- 1}}{1 + e^{- 1}}$
Commented by mathmax by abdo last updated on 28/Jul/19
$5) we have f(x) =∫_0 ^1 (dt/(x+e^t )) ⇒f^((n)) (x) =∫_0 ^1 (((−1)^n n!)/((x+e^t )^(n+1) ))dt =(−1)^n n! ∫_0 ^1 (dt/((x+e^t )^(n+1) )) 6) f(x) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n but f^((n)) (0) =(−1)^n n! ∫_0 ^1 (dt/e^((n+1)t) ) =(−1)^n n! ∫_0 ^1 e^(−(n+1)t) dt =(−1)^n n! [−(1/(n+1))e^(−(n+1)t) ]_0 ^1 =−(((−1)^n n!)/(n+1)){ e^(−(n+1)) −1} ⇒f(x) =Σ_(n=0) ^∞ (((−1)^(n+1) )/(n+1)){e^(−(n+1)) −1} ⇒f(x) =Σ_(n=0) ^∞ (((−1)^n )/(n+1)){1−e^(−(n+1)) } .$
$5) w e h a v e f (x) = \int_{0}^{1} \frac{d t}{x + e^{t}} \Rightarrow f^{(n)} (x) = \int_{0}^{1} \frac{{(- 1)}^{n} n!}{{(x + e^{t})}^{n + 1}} d t$ $= {(- 1)}^{n} n! \int_{0}^{1} \frac{d t}{{(x + e^{t})}^{n + 1}}$ $6) f (x) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} x^{n} b u t f^{(n)} (0) = {(- 1)}^{n} n! \int_{0}^{1} \frac{d t}{e^{(n + 1) t}}$ $= {(- 1)}^{n} n! \int_{0}^{1} e^{- (n + 1) t} d t = {(- 1)}^{n} n! {[- \frac{1}{n + 1} e^{- (n + 1) t}]}_{0}^{1}$ $= - \frac{{(- 1)}^{n} n!}{n + 1} {e^{- (n + 1)} - 1} \Rightarrow f (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n + 1}}{n + 1} {e^{- (n + 1)} - 1}$ $\Rightarrow f (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} {1 - e^{- (n + 1)}} .$
Commented by mathmax by abdo last updated on 28/Jul/19
$f(x) =Σ_(n=0) ^∞ (((−1)^n )/(n+1)){1−e^(−(n+1)) } x^n$
$f (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} {1 - e^{- (n + 1)}} x^{n}$
	Answered by Eminem last updated on 28/Jul/19
	$f(x)=∫_0 ^1 (e^(−t) /(xe^(−t) +1))dt=−(1/x)∫_0 ^1 ((−xe^(−t) )/(1+xe^(−t) ))dt ∀x∈IR^∗ f(x)=−(1/x){ln(1+xe^(−1) )−ln(1+x) }=(1/x)ln(((1+x)/(1+xe^(−1) ))) 2⟩f(2)=(1/2)ln((3/(1+2e^(−1) ))) 3)find g(x}=∫_0 ^1 (dt/((x+e^t )^2 )) ∀x∈IR^∗ f(x)=(1/x)ln(((1+x)/(1+xe^(−1) ))) f^′ (x)=∫_0 ^1 (d/dx)((1/(x+e^t )))dt=∫_0 ^1 ((−1)/((x+e^t )^2 ))dt=−g(x)==> g(x)=−f^′ (x)=−(d/dx){(1/x)ln(((1+x)/(1+xe^(−1) )))} =−{((−1)/x^2 )ln(((1+x)/(1+xe^(−1) )))+(1/x)×(((1+xe^(−1) −e^(−1) (1+x))/(((1+xe^(−1) )^2 )/(((1+x))/((1+xe^(−1) ))))))} =(1/x^2 )ln(((1+x)/(1+xe^(−1) )))−(1/x)×(((1−e^(−1) ))/((1+x)(1+xe^(−1) )))=g(x) ∫_0 ^1 (dt/((1+e^t )^2 ))=g(1)=ln((2/(1+e^(−1) )))−(((1−e^(−1) ))/((1+e^(−1) )(2))) f^n (x)=∫_0 ^1 (((−1)^n n!)/((x+e^t )^(n+1) )) f(x)=(1/x){ln(1+x)−ln(1+xe^(−1) )} =(1/x){Σ_(k=0) ((−1)^k (x^(k+1) /(k+1))−(−1)^k (((e^(−1) x)^(k+1) )/(k+1))) =Σ_(k=0) (−1)^k (1−e^(−(k+1)) )(x^k /(k+1))$
	$f (x) = \int_{0}^{1} \frac{e^{- t}}{{x e}^{- t} + 1} d t = - \frac{1}{x} \int_{0}^{1} \frac{- {x e}^{- t}}{1 + {x e}^{- t}} d t \forall x \in {I R}^{}$ $f (x) = - \frac{1}{x} {l n (1 + {x e}^{- 1}) - l n (1 + x)} = \frac{1}{x} l n (\frac{1 + x}{1 + {x e}^{- 1}})$ $2 ⟩ f (2) = \frac{1}{2} l n (\frac{3}{1 + 2 e^{- 1}})$ $3) f i n d g (x} = \int_{0}^{1} \frac{d t}{{(x + e^{t})}^{2}}$ $\forall x \in {I R}^{} f (x) = \frac{1}{x} l n (\frac{1 + x}{1 + {x e}^{- 1}})$ $f^{^{'}} (x) = \int_{0}^{1} \frac{d}{d x} (\frac{1}{x + e^{t}}) d t = \int_{0}^{1} \frac{- 1}{{(x + e^{t})}^{2}} d t = - g (x) ==>$ $g (x) = - f^{^{'}} (x) = - \frac{d}{d x} {\frac{1}{x} l n (\frac{1 + x}{1 + {x e}^{- 1}})}$ $= - {\frac{- 1}{x^{2}} l n (\frac{1 + x}{1 + {x e}^{- 1}}) + \frac{1}{x} \times (\frac{1 + {x e}^{- 1} - e^{- 1} (1 + x)}{\frac{{(1 + {x e}^{- 1})}^{2}}{\frac{(1 + x)}{(1 + {x e}^{- 1})}}})}$ $= \frac{1}{x^{2}} l n (\frac{1 + x}{1 + {x e}^{- 1}}) - \frac{1}{x} \times \frac{(1 - e^{- 1})}{(1 + x) (1 + {x e}^{- 1})} = g (x)$ $\int_{0}^{1} \frac{d t}{{(1 + e^{t})}^{2}} = g (1) = l n (\frac{2}{1 + e^{- 1}}) - \frac{(1 - e^{- 1})}{(1 + e^{- 1}) (2)}$ $f^{n} (x) = \int_{0}^{1} \frac{{(- 1)}^{n} n!}{{(x + e^{t})}^{n + 1}}$ $f (x) = \frac{1}{x} {l n (1 + x) - l n (1 + {x e}^{- 1})}$ $= \frac{1}{x} {\sum_{k = 0} ({(- 1)}^{k} \frac{x^{k + 1}}{k + 1} - {(- 1)}^{k} \frac{{(e^{- 1} x)}^{k + 1}}{k + 1})$ $= \sum_{k = 0} {(- 1)}^{k} (1 - e^{- (k + 1)}) \frac{x^{k}}{k + 1}$
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