1) find f(a) =∫_0 ^1 (dx/((ax+1)(√(x^2 −x+1)))) with a>0 2) calculate f^′ (a) 3)find the value of ∫_0 ^1 ((xdx)/((ax+1)^2 (√(x^2 −x+1)))) 4) calculate ∫_0 ^1 (dx/((2x+1)(√(x^2 −x+1)))) and ∫


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Question Number 53228 by maxmathsup by imad last updated on 19/Jan/19

$1) f i n d f (a) = \int_{0}^{1} \frac{d x}{(a x + 1) \sqrt{x^{2} - x + 1}} w i t h a > 0$ $2) c a l c u l a t e f^{^{'}} (a)$ $3) f i n d t h e v a l u e o f \int_{0}^{1} \frac{x d x}{{(a x + 1)}^{2} \sqrt{x^{2} - x + 1}}$ $4) c a l c u l a t e \int_{0}^{1} \frac{d x}{(2 x + 1) \sqrt{x^{2} - x + 1}} a n d \int_{0}^{1} \frac{x d x}{{(2 x + 1)}^{2} \sqrt{x^{2} - x + 1}}$
Commented by Abdo msup. last updated on 20/Jan/19
$1)we have x^2 −x +1 =(x−(1/2))^2 +(3/4) changement x−(1/2) =((√3)/2)sh(t) give f(a) = ∫_(argsh(−(1/(√3)))) ^(argsh((1/(√3)))) (1/((a((1/2)+((√3)/2)sh(t))+1)((√3)/2)ch(t)))((√3)/2)ch(t)dt =∫_(ln(−(1/(√3)) +(2/(√3)))) ^(ln((1/(√3)) +(2/(√3)))) ((2dt)/(a +a(√3)sh(t)+2)) =∫_(ln((1/(√3)))) ^(ln((√3))) ((2dt)/(a+a(√3)((e^t −e^(−t) )/2)+2)) = 4 ∫_(ln((1/(√3)))) ^(ln((√3))) (dt/(2a +a(√3)(e^t −e^(−t) )+4)) =_(e^t =u) 4 ∫_(1/(√3)) ^(√3) (1/(2a+a(√3)(u−u^(−1) )+4)) (du/u) =4 ∫_(1/(√3)) ^(√3) (du/(2au +a(√3)u^2 −a(√3) +4u)) =4 ∫_(1/(√3)) ^(√3) (du/(a(√3)u^2 +(2a+4)u−a(√3))) roots of p(u)=a(√3)u^2 +(2a+4)u−a(√3) Δ^′ =(a+2)^2 +3a^2 =a^2 +4a +4 +3a^2 =4a^2 +4a + u_1 =((−a−2+2(√(1+a+a^2 )))/(a(√3))) u_2 =((−a−2−2(√(1+a+a^2 )))/(a(√3))) ⇒ F(u) =(1/(a(√3)(u−u_1 )(u−u_2 ))) =(1/(a(√3)))(1/(u_1 −u_2 )){ (1/(u−u_1 )) −(1/(u−u_2 ))} =(1/(a(√3))) (1/((4(√(1+a+a^2 )))/(a(√3)))) {(1/(u−u_1 )) −(1/(u−u_2 ))} =(1/(4(√(1+a+a^2 )))){(1/(u−u_1 )) −(1/(u−u_2 ))}? ⇒ f(a) =(1/(√(1+a+a^2 ))) [ln∣((u−u_1 )/(u−u_2 ))∣]_(1/(√3)) ^(√3) =(1/(√(1+a+a^2 ))){ln∣(((√3)−u_1 )/((√3)−u_2 ))∣−ln∣(((1/(√3))−u_1 )/((1/(√3))−u_2 ))∣} =(1/(√(1+a+a^2 ))){ln∣(((√3)−((−a−2+2(√(1+a+a^2 )))/(a(√3))))/((√3)−((−a−2+2(√(1+a+a^2 )))/(a(√3)))))∣ −ln∣((1−(√3)((−a−2+2(√(1+a+a^2 )))/(a(√3))))/(1−(√3)((−a−2 −2(√(1+a+a^2 )))/(a(√3)))))∣ =(1/(√(1+a+a^2 ))){ ln∣((4a+2+2(√(1+a+a^2 )))/(4a+2−2(√(1+a+a^2 ))))∣ −ln∣ ((2a(√3)+2−2(√(1+a+a^2 )))/(2a(√3)+2 +2(√(1+a+a^2 ))))∣} ⇒f(a) =(1/(√(1+a+a^2 ))){ ln∣((2a+1+(√(1+a+a^2 )))/(2a+1−(√(1+a+a^2 ))))∣ −ln∣((a(√3)+1−(√(1+a+a^2 )))/(a(√3)+1 +(√(1+a+a^2 ))))∣} .$
$1) w e h a v e x^{2} - x + 1 = {(x - \frac{1}{2})}^{2} + \frac{3}{4} c h a n g e m e n t$ $x - \frac{1}{2} = \frac{\sqrt{3}}{2} s h (t) g i v e$ $f (a) = \int_{a r g s h (- \frac{1}{\sqrt{3}})}^{a r g s h (\frac{1}{\sqrt{3}})} \frac{1}{(a (\frac{1}{2} + \frac{\sqrt{3}}{2} s h (t)) + 1) \frac{\sqrt{3}}{2} c h (t)} \frac{\sqrt{3}}{2} c h (t) d t$ $= \int_{l n (- \frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})}^{l n (\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}})} \frac{2 d t}{a + a \sqrt{3} s h (t) + 2}$ $= \int_{l n (\frac{1}{\sqrt{3}})}^{l n (\sqrt{3})} \frac{2 d t}{a + a \sqrt{3} \frac{e^{t} - e^{- t}}{2} + 2}$ $= 4 \int_{l n (\frac{1}{\sqrt{3}})}^{l n (\sqrt{3})} \frac{d t}{2 a + a \sqrt{3} (e^{t} - e^{- t}) + 4}$ $=_{e^{t} = u} 4 \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{2 a + a \sqrt{3} (u - u^{- 1}) + 4} \frac{d u}{u}$ $= 4 \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{d u}{2 a u + a \sqrt{3} u^{2} - a \sqrt{3} + 4 u}$ $= 4 \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{d u}{a \sqrt{3} u^{2} + (2 a + 4) u - a \sqrt{3}}$ $r o o t s o f p (u) = a \sqrt{3} u^{2} + (2 a + 4) u - a \sqrt{3}$ $Δ^{^{'}} = {(a + 2)}^{2} + 3 a^{2} = a^{2} + 4 a + 4 + 3 a^{2} = 4 a^{2} + 4 a +$ $u_{1} = \frac{- a - 2 + 2 \sqrt{1 + a + a^{2}}}{a \sqrt{3}}$ $u_{2} = \frac{- a - 2 - 2 \sqrt{1 + a + a^{2}}}{a \sqrt{3}} \Rightarrow$ $F (u) = \frac{1}{a \sqrt{3} (u - u_{1}) (u - u_{2})}$ $= \frac{1}{a \sqrt{3}} \frac{1}{u_{1} - u_{2}} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}}$ $= \frac{1}{a \sqrt{3}} \frac{1}{\frac{4 \sqrt{1 + a + a^{2}}}{a \sqrt{3}}} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}}$ $= \frac{1}{4 \sqrt{1 + a + a^{2}}} {\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}} ? \Rightarrow$ $f (a) = \frac{1}{\sqrt{1 + a + a^{2}}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$ $= \frac{1}{\sqrt{1 + a + a^{2}}} {l n ∣ \frac{\sqrt{3} - u_{1}}{\sqrt{3} - u_{2}} ∣ - l n ∣ \frac{\frac{1}{\sqrt{3}} - u_{1}}{\frac{1}{\sqrt{3}} - u_{2}} ∣}$ $= \frac{1}{\sqrt{1 + a + a^{2}}} {l n ∣ \frac{\sqrt{3} - \frac{- a - 2 + 2 \sqrt{1 + a + a^{2}}}{a \sqrt{3}}}{\sqrt{3} - \frac{- a - 2 + 2 \sqrt{1 + a + a^{2}}}{a \sqrt{3}}} ∣$ $- l n ∣ \frac{1 - \sqrt{3} \frac{- a - 2 + 2 \sqrt{1 + a + a^{2}}}{a \sqrt{3}}}{1 - \sqrt{3} \frac{- a - 2 - 2 \sqrt{1 + a + a^{2}}}{a \sqrt{3}}} ∣$ $= \frac{1}{\sqrt{1 + a + a^{2}}} {l n ∣ \frac{4 a + 2 + 2 \sqrt{1 + a + a^{2}}}{4 a + 2 - 2 \sqrt{1 + a + a^{2}}} ∣$ $- l n ∣ \frac{2 a \sqrt{3} + 2 - 2 \sqrt{1 + a + a^{2}}}{2 a \sqrt{3} + 2 + 2 \sqrt{1 + a + a^{2}}} ∣}$ $\Rightarrow f (a) = \frac{1}{\sqrt{1 + a + a^{2}}} {l n ∣ \frac{2 a + 1 + \sqrt{1 + a + a^{2}}}{2 a + 1 - \sqrt{1 + a + a^{2}}} ∣$ $- l n ∣ \frac{a \sqrt{3} + 1 - \sqrt{1 + a + a^{2}}}{a \sqrt{3} + 1 + \sqrt{1 + a + a^{2}}} ∣} .$
Commented by Abdo msup. last updated on 20/Jan/19

$w e h a v e f^{^{'}} (a) = \int_{0}^{1} \frac{- x d x}{{(a x + 1)}^{2} \sqrt{x^{2} - x + 1}} \Rightarrow$ $\int_{0}^{1} \frac{x d x}{{(a x + 1)}^{2} \sqrt{x^{2} - x + 1}} = - f^{^{'}} (a) r e s t c a l c u l u s o f$ $f^{^{'}} (a) . . b e c o n t i n u e d . . .$
Commented by Abdo msup. last updated on 20/Jan/19
$4) ∫_0 ^1 (dx/((2x+1)(√(x^2 −x +1)))) =f(2) =(1/(√(1+2 +2^2 ))){ln∣((5+(√(1+2+2^2 )))/(5−(√(1+2+2^2 ))))∣−ln∣((2(√3)+1−(√(1+2+2^2 )))/(2(√3)+1+(√(1+2+2^2 ))))∣} =(1/(√7)){ln∣((5+(√7))/(5−(√7)))∣−ln∣((2(√3)+1−(√7))/(2(√3)+1+(√7)))∣} .$
$4) \int_{0}^{1} \frac{d x}{(2 x + 1) \sqrt{x^{2} - x + 1}} = f (2)$ $= \frac{1}{\sqrt{1 + 2 + 2^{2}}} {l n ∣ \frac{5 + \sqrt{1 + 2 + 2^{2}}}{5 - \sqrt{1 + 2 + 2^{2}}} ∣ - l n ∣ \frac{2 \sqrt{3} + 1 - \sqrt{1 + 2 + 2^{2}}}{2 \sqrt{3} + 1 + \sqrt{1 + 2 + 2^{2}}} ∣}$ $= \frac{1}{\sqrt{7}} {l n ∣ \frac{5 + \sqrt{7}}{5 - \sqrt{7}} ∣ - l n ∣ \frac{2 \sqrt{3} + 1 - \sqrt{7}}{2 \sqrt{3} + 1 + \sqrt{7}} ∣} .$
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