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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) 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 \dots

Commented by Abdo msup. last updated on 20/Jan/19

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}} ∣} .