let f(x)=∫_0 ^1 (dt/(t+x+(√(t^2 +1)))) (x real parametre) 1) find a explicite form forf(x) 2)detemine also g(x) =∫_0 ^1 (dt/((t+x+(√(t^2 +1)))^2 )) 3)give f^((n)) (x) at form of integrals 4) find the values of ∫_0 ^1 (dt/(t+(√(t^2 +1)))) and ∫_0 ^1 (dt/((t+(√(t^2 +1)))^2 )) 5) find the values of ∫_0 ^1 (dt/(t+1 +(√(t^2 +1)))) and ∫


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Question Number 64429 by mathmax by abdo last updated on 17/Jul/19

$l e t f (x) = \int_{0}^{1} \frac{d t}{t + x + \sqrt{t^{2} + 1}} (x r e a l p a r a m e t r e)$ $1) f i n d a e x p l i c i t e f o r m f o r f (x)$ $2) d e t e m i n e a l s o g (x) = \int_{0}^{1} \frac{d t}{{(t + x + \sqrt{t^{2} + 1})}^{2}}$ $3) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l s$ $4) f i n d t h e v a l u e s o f \int_{0}^{1} \frac{d t}{t + \sqrt{t^{2} + 1}} a n d \int_{0}^{1} \frac{d t}{{(t + \sqrt{t^{2} + 1})}^{2}}$ $5) f i n d t h e v a l u e s o f \int_{0}^{1} \frac{d t}{t + 1 + \sqrt{t^{2} + 1}} a n d \int_{0}^{1} \frac{d t}{{(t + 1 + \sqrt{t^{2} + 1})}^{2}}$
Commented by mathmax by abdo last updated on 20/Jul/19
$1) we have f(x) =∫_0 ^1 (dt/(t+x+(√(t^2 +1)))) changement t =sh(u) give f(x) =∫_0 ^(ln(1+(√2))) ((chu du)/(sh(u)+x +ch(u))) =∫_0 ^(ln(1+(√2))) (((e^u +e^(−u) )/2)/(((e^u −e^(−u) )/2)+x+((e^u +e^(−u) )/2)))du =∫_0 ^(ln(1+(√2))) ((e^u +e^(−u) )/(e^u −e^(−u) +2x +e^u +e^(−u) ))du =∫_0 ^(ln(1+(√2))) ((e^u +e^(−u) )/(2x +2e^u ))du =_(e^u =z) (1/2)∫_1 ^(1+(√2)) ((z +z^(−1) )/(x +z)) (dz/z) =(1/2) ∫_1 ^(1+(√2)) ((z+z^(−1) )/(xz +z^2 ))dz =(1/2) ∫_1 ^(1+(√2)) ((z^2 +1)/(z^2 (x+z)))dz let decompose F(z)=((z^2 +1)/(z^2 (x+z))) F(z) =(a/z) +(b/z^2 ) +(c/(x+z)) b =lim_(z→0) z^2 F(z) =(1/x) c =lim_(z→−x) (z+x)F(z) =((x^2 +1)/x^2 ) ⇒ F(z) =(a/z) +(1/(xz^2 )) +((x^2 +1)/(x^2 (z+x))) F(1) =(2/(x+1)) =a +(1/x) +((x^2 +1)/(x^2 (x+1))) ⇒2=(x+1)a+((x+1)/x) +((x^2 +1)/x^2 ) ⇒ 2=(x+1)a +((x^2 +x +x^2 +1)/x^2 ) =(x+1)a +((2x^2 +x+1)/x^2 ) ⇒ 2x^2 =x^2 (x+1)a +2x^2 +x+1 ⇒x^2 (x+1)a =−(x+1) ⇒ a =−(1/x^2 ) (if x≠−1) ⇒F(z) =−(1/(x^2 z)) +(1/(xz^2 )) +((x^2 +1)/(x^2 (z+x))) ⇒ 2f(x) =∫_1 ^(1+(√2)) {−(1/(x^2 z)) +(1/(xz^2 )) +((x^2 +1)/(x^2 (z+x)))}dz =−(1/x^2 ) ∫_1 ^(1+(√2)) (dz/z) +(1/x) ∫_1 ^(1+(√2)) (dz/z^2 ) +((x^2 +1)/x^2 ) ∫_1 ^(1+(√2)) (dz/(z+x)) =−(1/x^2 )[ln∣z∣]_1 ^(1+(√2)) −(1/x)[(1/z)]_1 ^(1+(√2)) +((x^2 +1)/x^2 )[ln∣z+x∣]_1 ^(1+(√2)) =((−ln(1+(√2)))/x^2 ) −(1/x)((1/(1+(√2))) −1) +((x^2 +1)/x^2 ){ln∣x+1+(√2)∣−ln∣x+1∣} ⇒ f(x)=(1/2){−((ln(1+(√2)))/x^2 ) +((√2)/x) +((x^2 +1)/x^2 )ln∣((x+1+(√2))/(x+1))∣ }$
$1) w e h a v e f (x) = \int_{0}^{1} \frac{d t}{t + x + \sqrt{t^{2} + 1}} c h a n g e m e n t t = s h (u) g i v e$ $f (x) = \int_{0}^{l n (1 + \sqrt{2})} \frac{c h u d u}{s h (u) + x + c h (u)}$ $= \int_{0}^{l n (1 + \sqrt{2})} \frac{\frac{e^{u} + e^{- u}}{2}}{\frac{e^{u} - e^{- u}}{2} + x + \frac{e^{u} + e^{- u}}{2}} d u$ $= \int_{0}^{l n (1 + \sqrt{2})} \frac{e^{u} + e^{- u}}{e^{u} - e^{- u} + 2 x + e^{u} + e^{- u}} d u = \int_{0}^{l n (1 + \sqrt{2})} \frac{e^{u} + e^{- u}}{2 x + 2 e^{u}} d u$ $=_{e^{u} = z} \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{z + z^{- 1}}{x + z} \frac{d z}{z} = \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{z + z^{- 1}}{x z + z^{2}} d z$ $= \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{z^{2} + 1}{z^{2} (x + z)} d z l e t d e c o m p o s e F (z) = \frac{z^{2} + 1}{z^{2} (x + z)}$ $F (z) = \frac{a}{z} + \frac{b}{z^{2}} + \frac{c}{x + z}$ $b = {l i m}_{z \to 0} z^{2} F (z) = \frac{1}{x}$ $c = {l i m}_{z \to - x} (z + x) F (z) = \frac{x^{2} + 1}{x^{2}} \Rightarrow$ $F (z) = \frac{a}{z} + \frac{1}{{x z}^{2}} + \frac{x^{2} + 1}{x^{2} (z + x)}$ $F (1) = \frac{2}{x + 1} = a + \frac{1}{x} + \frac{x^{2} + 1}{x^{2} (x + 1)} \Rightarrow 2 = (x + 1) a + \frac{x + 1}{x} + \frac{x^{2} + 1}{x^{2}} \Rightarrow$ $2 = (x + 1) a + \frac{x^{2} + x + x^{2} + 1}{x^{2}} = (x + 1) a + \frac{2 x^{2} + x + 1}{x^{2}} \Rightarrow$ $2 x^{2} = x^{2} (x + 1) a + 2 x^{2} + x + 1 \Rightarrow x^{2} (x + 1) a = - (x + 1) \Rightarrow$ $a = - \frac{1}{x^{2}} (i f x \neq - 1) \Rightarrow F (z) = - \frac{1}{x^{2} z} + \frac{1}{{x z}^{2}} + \frac{x^{2} + 1}{x^{2} (z + x)} \Rightarrow$ $2 f (x) = \int_{1}^{1 + \sqrt{2}} {- \frac{1}{x^{2} z} + \frac{1}{{x z}^{2}} + \frac{x^{2} + 1}{x^{2} (z + x)}} d z$ $= - \frac{1}{x^{2}} \int_{1}^{1 + \sqrt{2}} \frac{d z}{z} + \frac{1}{x} \int_{1}^{1 + \sqrt{2}} \frac{d z}{z^{2}} + \frac{x^{2} + 1}{x^{2}} \int_{1}^{1 + \sqrt{2}} \frac{d z}{z + x}$ $= - \frac{1}{x^{2}} {[l n ∣ z ∣]}_{1}^{1 + \sqrt{2}} - \frac{1}{x} {[\frac{1}{z}]}_{1}^{1 + \sqrt{2}} + \frac{x^{2} + 1}{x^{2}} {[l n ∣ z + x ∣]}_{1}^{1 + \sqrt{2}}$ $= \frac{- l n (1 + \sqrt{2})}{x^{2}} - \frac{1}{x} (\frac{1}{1 + \sqrt{2}} - 1) + \frac{x^{2} + 1}{x^{2}} {l n ∣ x + 1 + \sqrt{2} ∣ - l n ∣ x + 1 ∣} \Rightarrow$ $f (x) = \frac{1}{2} {- \frac{l n (1 + \sqrt{2})}{x^{2}} + \frac{\sqrt{2}}{x} + \frac{x^{2} + 1}{x^{2}} l n ∣ \frac{x + 1 + \sqrt{2}}{x + 1} ∣}$
Commented by mathmax by abdo last updated on 20/Jul/19

$2) t h e o r e m o f d e r i v a t i o n g i v e$ $f^{^{'}} (x) = \int_{0}^{1} \frac{\partial}{\partial x} (\frac{1}{t + x + \sqrt{t^{2} + 1}}) d t = - \int_{0}^{1} \frac{d t}{{(t + x + \sqrt{t^{2} + 1})}^{2}} = - g (x) \Rightarrow$ $g (x) = - f^{^{'}} (x) f i s k n o w n r e s t t o c a l c u l a t e f^{^{'}} (x)$
Commented by mathmax by abdo last updated on 20/Jul/19

$3) w e h a v e f^{(n)} (x) = \int_{0}^{1} \frac{\partial^{n}}{\partial x^{n}} (\frac{1}{t + x + \sqrt{t^{2} + 1}}) d t$ $= \int_{0}^{1} \frac{{(- 1)}^{n} n!}{{(t + x + \sqrt{t^{2} + 1})}^{n + 1}} d t$
Commented by mathmax by abdo last updated on 20/Jul/19
$4) ∫_0 ^1 (dt/(t+(√(t^2 +1)))) =_(t=sh(u)) ∫_0 ^(ln(1+(√2))) ((chu)/(sh(u)+ch(u)))du =∫_0 ^(ln(1+(√2))) ((e^u +e^(−u) )/(e^u −e^(−u) +e^u +e^(−u) )) du =∫_0 ^(ln(1+(√2))) ((e^u +e^(−u) )/(2e^u )) du =(1/2) ∫_0 ^(ln(1+(√2))) du +(1/2) ∫_0 ^(ln(1+(√2))) e^(−2u) du =((ln(1+(√2)))/2) −(1/4)[ e^(−2u) ]_0 ^(ln(1+(√2))) =((ln(1+(√2)))/2)−(1/4){ (1/((1+(√2))^2 )) −1} .$
$4) \int_{0}^{1} \frac{d t}{t + \sqrt{t^{2} + 1}} =_{t = s h (u)} \int_{0}^{l n (1 + \sqrt{2})} \frac{c h u}{s h (u) + c h (u)} d u$ $= \int_{0}^{l n (1 + \sqrt{2})} \frac{e^{u} + e^{- u}}{e^{u} - e^{- u} + e^{u} + e^{- u}} d u = \int_{0}^{l n (1 + \sqrt{2})} \frac{e^{u} + e^{- u}}{2 e^{u}} d u$ $= \frac{1}{2} \int_{0}^{l n (1 + \sqrt{2})} d u + \frac{1}{2} \int_{0}^{l n (1 + \sqrt{2})} e^{- 2 u} d u$ $= \frac{l n (1 + \sqrt{2})}{2} - \frac{1}{4} {[e^{- 2 u}]}_{0}^{l n (1 + \sqrt{2})}$ $= \frac{l n (1 + \sqrt{2})}{2} - \frac{1}{4} {\frac{1}{{(1 + \sqrt{2})}^{2}} - 1} .$
Commented by mathmax by abdo last updated on 20/Jul/19
$5) ∫_0 ^1 (dt/(t+1+(√(t^2 +1)))) =f(1) =(1/2){−ln(1+(√2))+(√2) +2ln∣((2+(√2))/2)∣$
$5) \int_{0}^{1} \frac{d t}{t + 1 + \sqrt{t^{2} + 1}} = f (1) = \frac{1}{2} {- l n (1 + \sqrt{2}) + \sqrt{2} + 2 l n ∣ \frac{2 + \sqrt{2}}{2} ∣$
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