let f(x) = ∫_0 ^1 (dt/(x +ch(t))) 1) find a explicite form of f(x) 2) calculate ∫_0 ^1 (dt/((x+ch(t))^2 )) 3) find the value of ∫_0 ^1 (dt/(1+ch(t))) and ∫_0 ^1 (dt/(2+ch(t))) 4) find the value of ∫_0 ^1 (dt/((1+cht)^2 )) and ∫


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Question Number 42487 by maxmathsup by imad last updated on 26/Aug/18

$l e t f (x) = \int_{0}^{1} \frac{d t}{x + c h (t)}$ $1) f i n d a e x p l i c i t e f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{1} \frac{d t}{{(x + c h (t))}^{2}}$ $3) f i n d t h e v a l u e o f \int_{0}^{1} \frac{d t}{1 + c h (t)} a n d \int_{0}^{1} \frac{d t}{2 + c h (t)}$ $4) f i n d t h e v a l u e o f \int_{0}^{1} \frac{d t}{{(1 + c h t)}^{2}} a n d \int_{0}^{1} \frac{d t}{{(2 + c h t)}^{2}}$
Commented by maxmathsup by imad last updated on 28/Aug/18
$1) we have f(x) = ∫_0 ^1 (dt/(x+((e^t +e^(−t) )/2))) =∫_0 ^1 ((2dt)/(2x +e^t +e^(−t) )) changement e^t =u give f(x) = ∫_1 ^e (2/(2x +u +(1/u))) (du/u) = ∫_1 ^e ((2du)/(2xu +u^2 +1)) = ∫_1 ^x ((2du)/(u^2 +2xu +1)) Δ^′ =x^2 −1 case 1 ∣x∣>1 ⇒u_1 =−x+(√(x^2 −1)) and u_2 =−x−(√(x^2 −1)) ⇒ F(u) = (2/(u^2 +2xu +1)) = (a/(u−u_1 )) +(b/(u−u_2 )) (=(2/((u−u_1 )(u−u_2 )))) a = (2/(u_1 −u_2 )) =(2/(2(√(x^2 −1)))) =(1/(√(x^2 −1))) b =(2/(u_2 −u_1 )) =(2/(−2(√(x^2 −1)))) =−(1/(√(x^2 −1))) ⇒ f(x) = ∫_1 ^e F(u)du =(1/(√(x^2 −1))) ∫_1 ^e ( (1/(u−u_1 )) −(1/(u−u_2 )))du =(1/(√(x^2 −1)))[ln∣((u−u_1 )/(u−u_2 ))∣]_1 ^e =(1/(√(x^2 −1))){ln∣ ((e+x−(√(x^2 −1)))/(e+x+(√(x^2 −1))))∣ −ln∣((1+x−(√(x^2 −1)))/(1+x +(√(x^2 −1))))∣} case 2 ∣x∣<1 ⇒f(x)= ∫_1 ^e ((2du)/(u^2 +2xu +x^2 +1−x^2 )) =∫_1 ^e ((2du)/((u+x)^2 +1−x^2 )) =_(u+x =(√(1−x^2 ))α) ∫_((1+x)/(√(1−x^2 ))) ^((e+x)/(√(1−x^2 ))) ((2(√(1−x^2 ))dα)/((1−x^2 )(1+α^2 ))) =(2/(√(1−x^2 ))) [ arctan(α)]_((1+x)/(√(1−x^2 ))) ^((e+x)/(√(1−x^2 ))) =(2/(√(1−x^2 ))){ arctan(((e+x)/(√(1−x^2 ))))−arctan(((1+x)/(√(1−x^2 ))))} .$
$1) w e h a v e f (x) = \int_{0}^{1} \frac{d t}{x + \frac{e^{t} + e^{- t}}{2}} = \int_{0}^{1} \frac{2 d t}{2 x + e^{t} + e^{- t}} c h a n g e m e n t e^{t} = u g i v e$ $f (x) = \int_{1}^{e} \frac{2}{2 x + u + \frac{1}{u}} \frac{d u}{u} = \int_{1}^{e} \frac{2 d u}{2 x u + u^{2} + 1} = \int_{1}^{x} \frac{2 d u}{u^{2} + 2 x u + 1}$ $Δ^{^{'}} = x^{2} - 1 c a s e 1 ∣ x ∣> 1 \Rightarrow u_{1} = - x + \sqrt{x^{2} - 1} a n d u_{2} = - x - \sqrt{x^{2} - 1} \Rightarrow$ $F (u) = \frac{2}{u^{2} + 2 x u + 1} = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} (= \frac{2}{(u - u_{1}) (u - u_{2})})$ $a = \frac{2}{u_{1} - u_{2}} = \frac{2}{2 \sqrt{x^{2} - 1}} = \frac{1}{\sqrt{x^{2} - 1}}$ $b = \frac{2}{u_{2} - u_{1}} = \frac{2}{- 2 \sqrt{x^{2} - 1}} = - \frac{1}{\sqrt{x^{2} - 1}} \Rightarrow$ $f (x) = \int_{1}^{e} F (u) d u = \frac{1}{\sqrt{x^{2} - 1}} \int_{1}^{e} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) d u$ $= \frac{1}{\sqrt{x^{2} - 1}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{1}^{e} = \frac{1}{\sqrt{x^{2} - 1}} {l n ∣ \frac{e + x - \sqrt{x^{2} - 1}}{e + x + \sqrt{x^{2} - 1}} ∣ - l n ∣ \frac{1 + x - \sqrt{x^{2} - 1}}{1 + x + \sqrt{x^{2} - 1}} ∣}$ $c a s e 2 ∣ x ∣< 1 \Rightarrow f (x) = \int_{1}^{e} \frac{2 d u}{u^{2} + 2 x u + x^{2} + 1 - x^{2}} = \int_{1}^{e} \frac{2 d u}{{(u + x)}^{2} + 1 - x^{2}}$ $=_{u + x = \sqrt{1 - x^{2}} α} \int_{\frac{1 + x}{\sqrt{1 - x^{2}}}}^{\frac{e + x}{\sqrt{1 - x^{2}}}} \frac{2 \sqrt{1 - x^{2}} d α}{(1 - x^{2}) (1 + α^{2})}$ $= \frac{2}{\sqrt{1 - x^{2}}} {[a r c t a n (α)]}_{\frac{1 + x}{\sqrt{1 - x^{2}}}}^{\frac{e + x}{\sqrt{1 - x^{2}}}} = \frac{2}{\sqrt{1 - x^{2}}} {a r c t a n (\frac{e + x}{\sqrt{1 - x^{2}}}) - a r c t a n (\frac{1 + x}{\sqrt{1 - x^{2}}})} .$
Commented by maxmathsup by imad last updated on 28/Aug/18

$2) w e h a v e f^{^{'}} (x) = \int_{0}^{1} \frac{\partial}{\partial x} (\frac{1}{x + c h (t)}) d t = - \int_{0}^{1} \frac{d t}{{(x + c h (t))}^{2}} \Rightarrow$ $\int_{0}^{1} \frac{d t}{{(x + c h (t))}^{2}} = - f^{^{'}} (x) a n d f i s k n o w n . .$
Commented by maxmathsup by imad last updated on 28/Aug/18
$3) ∫_0 ^1 (dt/(2+ch(t))) =f(2) = (1/(√3)){ln∣((e+2−(√3))/(e+2+(√3)))∣ −ln∣ ((3−(√3))/(3+(√3)))∣ } .$
$3) \int_{0}^{1} \frac{d t}{2 + c h (t)} = f (2) = \frac{1}{\sqrt{3}} {l n ∣ \frac{e + 2 - \sqrt{3}}{e + 2 + \sqrt{3}} ∣ - l n ∣ \frac{3 - \sqrt{3}}{3 + \sqrt{3}} ∣} .$
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