let f(x) =∫_0 ^1 (dt/(1+x(√(1+t^2 )))) with x>0 1)detemine a explicit form of f(x) 2)find also g(x) =∫_0 ^1 ((√(1+t^2 ))/((1+x(√(1+t^2 )))^2 ))dt 3) find the value of integrals ∫_0 ^1 (dt/(1+2(√(1+t^2 )))) and ∫


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Question Number 65773 by mathmax by abdo last updated on 03/Aug/19

$l e t f (x) = \int_{0}^{1} \frac{d t}{1 + x \sqrt{1 + t^{2}}} w i t h x > 0$ $1) d e t e m i n e a e x p l i c i t f o r m o f f (x)$ $2) f i n d a l s o g (x) = \int_{0}^{1} \frac{\sqrt{1 + t^{2}}}{{(1 + x \sqrt{1 + t^{2}})}^{2}} d t$ $3) f i n d t h e v a l u e o f i n t e g r a l s \int_{0}^{1} \frac{d t}{1 + 2 \sqrt{1 + t^{2}}} a n d$ $\int_{0}^{1} \frac{d t}{{(1 + 2 \sqrt{1 + t^{2}})}^{2}}$
Commented by mathmax by abdo last updated on 06/Aug/19
$2)we have f^′ (x) =−∫_0 ^1 (((√(1+t^2 ))dt)/((1+x(√(1+t^2 )))^2 )) =−g(x) ⇒g(x)=−f^′ (x) rest to calculate f^′ (x) 3)changement t =sht give I=∫_0 ^1 (dt/(1+2(√(1+t^2 )))) =∫_0 ^(ln(1+(√2))) ((ch(t)dt)/(1+2ch(t))) =∫_0 ^(ln(1+(√2))) (((e^t +e^(−t) )/2)/(1+2((e^t +e^(−t) )/2)))dt =(1/2)∫_0 ^(ln(1+(√2))) ((e^t +e^(−t) )/(1+e^t +e^(−t) ))dt =_(e^t =x) (1/2)∫_1 ^(1+(√2)) ((x+x^(−1) )/(1+x+x^(−1) ))(dx/x) =(1/2) ∫_1 ^(1+(√2)) ((x^2 +1)/(x^2 (1+x+x^(−1) )))dx =(1/2)∫_1 ^(1+(√2)) ((x^2 +1)/(x^2 +x^3 +x))dx =(1/2)∫_1 ^(1+(√2)) ((x^2 +1)/(x(x^2 +x+1)))dx let decomposeF(x)=((x^2 +1)/(x(x^2 +x+1))) F(x) =(a/x) +((bx+c)/(x^2 +x+1)) a =lim_(x→0) xF(x) =1 lim_(x→+∞) xF(x) =1 =a+b ⇒b=1−a=0 ⇒F(x)=(1/x) +(c/(x^2 +x+1)) F(1) =(2/3) =1 +(c/3) ⇒2 =3 +c ⇒c =−1⇒F(x)=(1/x)−(1/(x^2 +x+1)) ⇒ I =(1/2){ ∫_1 ^(1+(√2)) (dx/x)−∫_1 ^(1+(√2)) (dx/(x^2 +x+1))} =(1/2)ln(1+(√2))−(1/2)∫_1 ^(1+(√2)) (dx/(x^2 +x+1)) ∫_1 ^(1+(√2)) (dx/(x^2 +x+1)) =∫_1 ^(1+(√2)) (dx/((x+(1/2))^2 +(3/4))) =_(x+(1/2)=((√3)/2)u) (4/3) ∫_(√3) ^((3+2(√2))/3) (1/(1+u^2 ))((√3)/2)du = (2/(√3)) [arctanu]_(√3) ^((3+2(√2))/3) =(2/(√3)){ arctan(((3+2(√2))/3))−arctan((√3))}⇒ I =(1/2)ln(1+(√2))−(1/(√3)){arctan(((3+2(√2))/3))−arctan((√3))}$
$2) w e h a v e f^{^{'}} (x) = - \int_{0}^{1} \frac{\sqrt{1 + t^{2}} d t}{{(1 + x \sqrt{1 + t^{2}})}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x)$ $r e s t t o c a l c u l a t e f^{^{'}} (x)$ $3) c h a n g e m e n t t = s h t g i v e I = \int_{0}^{1} \frac{d t}{1 + 2 \sqrt{1 + t^{2}}}$ $= \int_{0}^{l n (1 + \sqrt{2})} \frac{c h (t) d t}{1 + 2 c h (t)} = \int_{0}^{l n (1 + \sqrt{2})} \frac{\frac{e^{t} + e^{- t}}{2}}{1 + 2 \frac{e^{t} + e^{- t}}{2}} d t$ $= \frac{1}{2} \int_{0}^{l n (1 + \sqrt{2})} \frac{e^{t} + e^{- t}}{1 + e^{t} + e^{- t}} d t =_{e^{t} = x} \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{x + x^{- 1}}{1 + x + x^{- 1}} \frac{d x}{x}$ $= \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{x^{2} + 1}{x^{2} (1 + x + x^{- 1})} d x = \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{x^{2} + 1}{x^{2} + x^{3} + x} d x$ $= \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{x^{2} + 1}{x (x^{2} + x + 1)} d x l e t d e c o m p o s e F (x) = \frac{x^{2} + 1}{x (x^{2} + x + 1)}$ $F (x) = \frac{a}{x} + \frac{b x + c}{x^{2} + x + 1}$ $a = {l i m}_{x \to 0} x F (x) = 1$ ${l i m}_{x \to + \infty} x F (x) = 1 = a + b \Rightarrow b = 1 - a = 0 \Rightarrow F (x) = \frac{1}{x} + \frac{c}{x^{2} + x + 1}$ $F (1) = \frac{2}{3} = 1 + \frac{c}{3} \Rightarrow 2 = 3 + c \Rightarrow c = - 1 \Rightarrow F (x) = \frac{1}{x} - \frac{1}{x^{2} + x + 1}$ $\Rightarrow I = \frac{1}{2} {\int_{1}^{1 + \sqrt{2}} \frac{d x}{x} - \int_{1}^{1 + \sqrt{2}} \frac{d x}{x^{2} + x + 1}}$ $= \frac{1}{2} l n (1 + \sqrt{2}) - \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{d x}{x^{2} + x + 1}$ $\int_{1}^{1 + \sqrt{2}} \frac{d x}{x^{2} + x + 1} = \int_{1}^{1 + \sqrt{2}} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{\sqrt{3}}^{\frac{3 + 2 \sqrt{2}}{3}} \frac{1}{1 + u^{2}} \frac{\sqrt{3}}{2} d u$ $= \frac{2}{\sqrt{3}} {[a r c t a n u]}_{\sqrt{3}}^{\frac{3 + 2 \sqrt{2}}{3}} = \frac{2}{\sqrt{3}} {a r c t a n (\frac{3 + 2 \sqrt{2}}{3}) - a r c t a n (\sqrt{3})} \Rightarrow$ $I = \frac{1}{2} l n (1 + \sqrt{2}) - \frac{1}{\sqrt{3}} {a r c t a n (\frac{3 + 2 \sqrt{2}}{3}) - a r c t a n (\sqrt{3})}$
Commented by mathmax by abdo last updated on 06/Aug/19
$1)f(x) =∫_0 ^1 (dt/(1+x(√(1+t^2 )))) changement t =sh(u) give f(x)=∫_0 ^(ln(1+(√2))) ((chu)/(1+xchu))du =∫_0 ^(ln(1+(√2))) (((e^u +e^(−u) )/2)/(1+x((e^u +e^(−u) )/2)))du = ∫_0 ^(ln(1+(√2))) ((e^u +e^(−u) )/(2 +xe^u +x e^(−u) )) du =_(e^u =z) ∫_1 ^(1+(√2)) ((z+z^(−1) )/(2+xz+xz^(−1) ))(dz/z) = ∫_1 ^(1+(√2)) ((z^2 +1)/(z^2 (2+xz +xz^(−1) )))dz =∫_1 ^(1+(√2)) ((z^2 +1)/(2z^2 +xz^3 +xz))dz =∫_1 ^(1+(√2)) ((z^2 +1)/(z(xz^2 +2z+x)))dz let decomposeF(z) =((z^2 +1)/(z(xz^2 +2z+x))) xz^2 +2z +x =0 →Δ^′ =1−x^2 case 1 1−x^2 >0 ⇒0<x<1 ⇒z_1 =((−1+(√(1−x^2 )))/x) z_2 =((−1−(√(1−x^2 )))/x) ⇒F(z) =((z^2 +1)/(xz(z−z_1 )(z−z_2 ))) =(a/z) +(b/(z−z_1 )) +(c/(z−z_2 )) a =lim_(z→0) zF(z) =(1/x) b =lim_(z→z_1 ) (z−z_1 )F(z) =((z_1 ^2 +1)/(xz_1 (z_1 −z_2 ))) c =lim_(z→z_2 ) (z−z_2 )F(z) =((z_2 ^2 +1)/(xz_2 (z_2 −z_1 ))) ⇒ f(x) =∫_1 ^(1+(√2)) F(z)dz =[aln∣z∣+bln∣z−z_1 ∣+cln∣z−z_2 ∣]_1 ^(1+(√2)) =aln(1+(√2))+bln∣1+(√2)−z_1 ∣+cln∣1+(√2)−z_2 ∣−bln∣1−z_1 ∣ −cln∣1−z_2 ∣ case 2 1−x^2 <0 ⇒x>1 f(x)=∫_1 ^(1+(√2)) ((z^2 +1)/(z(xz^2 +2z +x)))dz F(z) =(a/z) +((bz +c)/(xz^2 +2z +x)) a =(1/x) lim_(z→+∞) zF(z) =(1/x) =a+(b/x) ⇒1 =ax +b ⇒b=1−ax=0 ⇒ F(z) =(1/(xz)) +(c/(xz^2 +2z +x)) F(1) =(2/((2x+2))) =(1/(x+1)) =(1/x) +(c/(2x+2)) ⇒1 =((x+1)/x) +(c/2) ⇒ (c/2) =−(1/x) ⇒c =((−2)/x) ⇒ F(z) =(1/(xz))−(2/x)(1/((xz^2 +2z +x))) =(1/x){ (1/z)−(2/(xz^2 +2z +x))} ⇒f(x) =(1/x){ ∫_1 ^(1+(√2)) (dz/z)−2∫_1 ^(1+(√2)) (dz/(xz^2 +2z+x))} =(1/x)ln(1+(√2))−(2/x) ∫_1 ^(1+(√2)) (dz/(xz^2 +2z +x)) =....$
$1) f (x) = \int_{0}^{1} \frac{d t}{1 + x \sqrt{1 + t^{2}}} c h a n g e m e n t t = s h (u) g i v e$ $f (x) = \int_{0}^{l n (1 + \sqrt{2})} \frac{c h u}{1 + x c h u} d u = \int_{0}^{l n (1 + \sqrt{2})} \frac{\frac{e^{u} + e^{- u}}{2}}{1 + x \frac{e^{u} + e^{- u}}{2}} d u$ $= \int_{0}^{l n (1 + \sqrt{2})} \frac{e^{u} + e^{- u}}{2 + {x e}^{u} + x e^{- u}} d u =_{e^{u} = z} \int_{1}^{1 + \sqrt{2}} \frac{z + z^{- 1}}{2 + x z + {x z}^{- 1}} \frac{d z}{z}$ $= \int_{1}^{1 + \sqrt{2}} \frac{z^{2} + 1}{z^{2} (2 + x z + {x z}^{- 1})} d z = \int_{1}^{1 + \sqrt{2}} \frac{z^{2} + 1}{2 z^{2} + {x z}^{3} + x z} d z$ $= \int_{1}^{1 + \sqrt{2}} \frac{z^{2} + 1}{z ({x z}^{2} + 2 z + x)} d z l e t d e c o m p o s e F (z) = \frac{z^{2} + 1}{z ({x z}^{2} + 2 z + x)}$ ${x z}^{2} + 2 z + x = 0 \to Δ^{^{'}} = 1 - x^{2}$ $c a s e 1 1 - x^{2} > 0 \Rightarrow 0 < x < 1 \Rightarrow z_{1} = \frac{- 1 + \sqrt{1 - x^{2}}}{x}$ $z_{2} = \frac{- 1 - \sqrt{1 - x^{2}}}{x} \Rightarrow F (z) = \frac{z^{2} + 1}{x z (z - z_{1}) (z - z_{2})}$ $= \frac{a}{z} + \frac{b}{z - z_{1}} + \frac{c}{z - z_{2}}$ $a = {l i m}_{z \to 0} z F (z) = \frac{1}{x}$ $b = {l i m}_{z \to z_{1}} (z - z_{1}) F (z) = \frac{z_{1}^{2} + 1}{{x z}_{1} (z_{1} - z_{2})}$ $c = {l i m}_{z \to z_{2}} (z - z_{2}) F (z) = \frac{z_{2}^{2} + 1}{{x z}_{2} (z_{2} - z_{1})} \Rightarrow$ $f (x) = \int_{1}^{1 + \sqrt{2}} F (z) d z = {[a l n ∣ z ∣ + b l n ∣ z - z_{1} ∣ + c l n ∣ z - z_{2} ∣]}_{1}^{1 + \sqrt{2}}$ $= a l n (1 + \sqrt{2}) + b l n ∣ 1 + \sqrt{2} - z_{1} ∣ + c l n ∣ 1 + \sqrt{2} - z_{2} ∣ - b l n ∣ 1 - z_{1} ∣$ $- c l n ∣ 1 - z_{2} ∣$ $c a s e 2 1 - x^{2} < 0 \Rightarrow x > 1 f (x) = \int_{1}^{1 + \sqrt{2}} \frac{z^{2} + 1}{z ({x z}^{2} + 2 z + x)} d z$ $F (z) = \frac{a}{z} + \frac{b z + c}{{x z}^{2} + 2 z + x}$ $a = \frac{1}{x}$ ${l i m}_{z \to + \infty} z F (z) = \frac{1}{x} = a + \frac{b}{x} \Rightarrow 1 = a x + b \Rightarrow b = 1 - a x = 0 \Rightarrow$ $F (z) = \frac{1}{x z} + \frac{c}{{x z}^{2} + 2 z + x}$ $F (1) = \frac{2}{(2 x + 2)} = \frac{1}{x + 1} = \frac{1}{x} + \frac{c}{2 x + 2} \Rightarrow 1 = \frac{x + 1}{x} + \frac{c}{2} \Rightarrow$ $\frac{c}{2} = - \frac{1}{x} \Rightarrow c = \frac{- 2}{x} \Rightarrow F (z) = \frac{1}{x z} - \frac{2}{x} \frac{1}{({x z}^{2} + 2 z + x)}$ $= \frac{1}{x} {\frac{1}{z} - \frac{2}{{x z}^{2} + 2 z + x}} \Rightarrow f (x) = \frac{1}{x} {\int_{1}^{1 + \sqrt{2}} \frac{d z}{z} - 2 \int_{1}^{1 + \sqrt{2}} \frac{d z}{{x z}^{2} + 2 z + x}}$ $= \frac{1}{x} l n (1 + \sqrt{2}) - \frac{2}{x} \int_{1}^{1 + \sqrt{2}} \frac{d z}{{x z}^{2} + 2 z + x} = . . . .$
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