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Question Number 44179 by abdo.msup.com last updated on 22/Sep/18

1) f i n d \int \frac{d x}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} - x + 1}}

2) c a l c u l a t e \int_{0}^{1} \frac{d x}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} - x + 1}}

Commented by maxmathsup by imad last updated on 07/Oct/18

l e t A = \int \frac{d x}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} - x + 1}} \Rightarrow A = \int \frac{\sqrt{x^{2} + x + 1} - \sqrt{x^{2} - x + 1}}{x^{2} + x + 1 - x^{2} + x - 1} d x

= \frac{1}{2} \int \frac{\sqrt{x^{2} + x + 1}}{x} d x - \frac{1}{2} \int \frac{\sqrt{x^{2} - x + 1}}{x} d x \Rightarrow 2 A = H - K

H = \int \frac{x^{2} + x + 1}{x \sqrt{x^{2} + x + 1}} d x = \int \frac{x}{\sqrt{x^{2} + x + 1}} d x + \int \frac{d x}{\sqrt{x^{2} + x + 1}} + \int \frac{d x}{x \sqrt{x^{2} + x + 1}}

b u t \int \frac{x}{\sqrt{x^{2} + x + 1}} d x = \frac{1}{2} \int \frac{2 x + 1 - 1}{\sqrt{x^{2} + x + 1}} d x

= \sqrt{x^{2} + x + 1} - \frac{1}{2} \int \frac{d x}{\sqrt{x^{2} + x + 1}} \Rightarrow H = \sqrt{x^{2} + x + 1} + \frac{1}{2} \int \frac{d x}{\sqrt{x^{2} + x + 1}} + \int \frac{d x}{x \sqrt{x^{2} + x + 1}}

\int \frac{d x}{\sqrt{x^{2} + x + 1}} = \int \frac{d x}{\sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t a n (t)} \frac{\sqrt{3}}{2} \int \frac{1 + {t a n}^{2} t}{\frac{\sqrt{3}}{2} \sqrt{1 + {t a n}^{2} t}} d t

= \int \sqrt{1 + {t a n}^{2} t} d t = \int \frac{d t}{c o s t} =_{t a n (\frac{t}{2}) = u} \int \frac{1}{\frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int \frac{2 d u}{1 - u^{2}} = \int (\frac{1}{1 - u} + \frac{1}{1 + u}) d u = l n ∣ \frac{1 + u}{1 - u} ∣ = l n ∣ \frac{1 + t a n (\frac{t}{2})}{1 - t a n (\frac{t}{2})} ∣

= l n ∣ t a n (\frac{π}{4} + \frac{t}{2}) ∣ = l n ∣ t a n (\frac{1}{2} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + \frac{π}{4}) ∣

\int \frac{d x}{x \sqrt{x^{2} + x + 1}} = \int \frac{d x}{x \sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} s h (t)} \frac{\sqrt{3}}{2} \int \frac{c h (t)}{(\frac{\sqrt{3}}{2} s h (t) - \frac{1}{2}) \frac{\sqrt{3}}{2} c h (t)} d t

= \int \frac{2 d t}{\sqrt{3} \frac{e^{t} - e^{- t}}{2} - 1} = \int \frac{4 d t}{\sqrt{3} e^{t} - \sqrt{3} e^{- t} - 2}

=_{e^{t} = u} \int \frac{4}{\sqrt{3} u - \sqrt{3} u^{- 1} - 2} \frac{d u}{u}

= \int \frac{4 d u}{\sqrt{3} u^{2} - \sqrt{3} - 2 u} d u = \int \frac{4 d u}{\sqrt{3} u^{2} - 2 u - \sqrt{3}}

Δ^{^{'}} = 1 + 3 = 4 \Rightarrow u_{1} = \frac{1 + 2}{\sqrt{3}} = \sqrt{3} a n d u_{2} = \frac{1 - 2}{\sqrt{3}} = - \frac{1}{\sqrt{3}}

\int F (u) d u = \int \frac{4 d u}{\sqrt{3} (u - \sqrt{3}) (u + \frac{1}{\sqrt{3}})} = \frac{4}{\sqrt{3}} \frac{1}{\sqrt{3} + \frac{1}{\sqrt{3}}} \int (\frac{1}{u - \sqrt{3}} - \frac{1}{u + \frac{1}{\sqrt{3}}}) d u

= l n ∣ \frac{u - \sqrt{3}}{u + \frac{1}{\sqrt{3}}} ∣ = l n ∣ \frac{e^{t} - \sqrt{3}}{e^{t} + \frac{1}{\sqrt{3}}} ∣ b u t t = a r g s h (\frac{2 x + 1}{\sqrt{3}}) = l n (\frac{2 x + 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}}) \Rightarrow

\int F (u) d u = l n ∣ \frac{\frac{2 x + 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}} - \sqrt{3}}{\frac{2 x + 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}} + \frac{1}{\sqrt{3}}} ∣ + c \Rightarrow

H = \sqrt{x^{2} + x + 1} + \frac{1}{2} l n ∣ t a n (\frac{1}{2} a r c t a n (\frac{2 x + 1}{\sqrt{3}}) + \frac{π}{4}) ∣

+ l n ∣ \frac{\frac{2 x + 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}} - \sqrt{3}}{\frac{2 x + 1}{\sqrt{3}} + \sqrt{1 + {(\frac{2 x + 1}{\sqrt{3}})}^{2}} + \frac{1}{\sqrt{3}}} ∣ + c

f o r d e t e r m i n a t i o n o f K w e u s e t h e c h a n g e m e n t x = - t \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18

\int \frac{d x}{\sqrt{x^{2} + x + 1} + \sqrt{x^{2} - x + 1}}

\int \frac{\sqrt{x^{2} + x + 1} - \sqrt{x^{2} - x + 1}}{(x^{2} + x + 1) - (x^{2} - x + 1)} d x

\int \frac{\sqrt{x^{2} + x + 1}}{2 x} d x - \int \frac{\sqrt{x^{2} - x + 1}}{2 x} d x

\frac{I_{1}}{2} - \frac{I_{2}}{2} = I

I_{1} = \int \frac{x^{2} + x + 1}{x \sqrt{x^{2} + x + 1}} d x

= \int \frac{x}{\sqrt{x^{2} + x + 1}} d x + \int \frac{d x}{\sqrt{x^{2} + x + 1}} + \int \frac{d x}{x \sqrt{x^{2} + x + 1}}

★ \frac{1}{2} \int \frac{2 x + 1 - 1}{\sqrt{x^{2} + x + 1}} d x + ★ ★ \int \frac{d x}{\sqrt{x^{2} + 2 . x . \frac{1}{2} + \frac{1}{4} + 1 - \frac{1}{4}}} + ★ ★ ★ t h i s p a r t s o l v i n g s e p a r T e l y

★ ★ ★ \int \frac{d x}{x \sqrt{x^{2} + x + 1}} x = \frac{1}{t} d x = - \frac{1}{t^{2}} d t

\int \frac{- d t}{t^{2} \times \frac{1}{t} \times \sqrt{\frac{1}{t^{2}} + \frac{1}{t} + 1}}

\int \frac{- t \times d t}{t \sqrt{t^{2} + t + 1}}

- \int \frac{d t}{\sqrt{{(t + \frac{1}{2})}^{2} + \frac{3}{4}}} = - l n ∣ (t + \frac{1}{2}) + \sqrt{{(t + \frac{1}{2})}^{2} + \frac{3}{4}} ∣

= - l n ∣ (\frac{1}{x} + \frac{1}{2}) + \sqrt{{(\frac{1}{x} + \frac{1}{2})}^{2} + \frac{3}{4}} ∣

n o w ★

\frac{1}{2} \int \frac{2 x + 1 - 1}{\sqrt{x^{2} + x + 1}}

\frac{1}{2} \int \frac{d (x^{2} + x + 1)}{\sqrt{x^{2} + x + 1}} - \frac{1}{2} \int \frac{d x}{\sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}}

\frac{1}{2} \times \frac{\sqrt{x^{2} + x + 1}}{\frac{1}{2}} - \frac{1}{2} l n ∣ (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} ∣

= \sqrt{x^{2} + x + 1} - \frac{1}{2} l n ∣ (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} ∣

n o w ★ ★ \int \frac{d x}{\sqrt{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}}

= l n ∣ (x + \frac{1}{2}) + \sqrt{x^{2} + x + 1} ∣

s i m i l a r l y t h e n x t I_{2} c a n b e s o l v e d . . l e n t h y q u e s t i o n \dots

\int \frac{\sqrt{x^{2} - x + 1}}{x} d x

\int \frac{x^{2} - x + 1}{x \sqrt{x^{2} - x + 1}} d x

★ \int \frac{x}{\sqrt{x^{2} - x + 1}} d x - ★ ★ \int \frac{d x}{\sqrt{x^{2} - x + 1}} + ★ ★ ★ \int \frac{d x}{x \sqrt{x^{2} - x + 1}}

★

\frac{1}{2} \int \frac{2 x - 1 + 1}{\sqrt{x^{2} - x + 1}} d x

\frac{1}{2} \int \frac{d (x^{2} - x + 1)}{\sqrt{x^{2} - x + 1}} d x + \frac{1}{2} \int \frac{d x}{\sqrt{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}}

\frac{1}{2} \times \frac{\sqrt{x^{2} - x + 1}}{\frac{1}{2}} + \frac{1}{2} l n ∣ (x - \frac{1}{2}) + \sqrt{x^{2} - x + 1} ∣

★ [★

\int \frac{d x}{\sqrt{x^{2} - x + 1}}

\int \frac{d x}{\sqrt{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}}

= l n ∣ (x - \frac{1}{2}) + \sqrt{x^{2} - x + 1} ∣

★ ★ ★

\int \frac{d x}{x \sqrt{x^{2} - x + 1}}

t = \frac{1}{x}

- 1 \int \frac{d t}{t^{2}} \times \frac{1}{\frac{1}{t} \sqrt{\frac{1}{t^{2}} - \frac{1}{t} + 1}}

- \int \frac{d t}{\sqrt{1 - t + t^{2}}}

- \int \frac{d t}{\sqrt{{(t - \frac{1}{2})}^{2} + \frac{3}{4}}}

- l n ∣ (t - \frac{1}{2}) + \sqrt{t^{2} - t + 1} ∣

- l n ∣ (\frac{1}{x} - \frac{1}{2}) + \sqrt{\frac{1}{x^{2}} - \frac{1}{x} + 1}

n o w p l s a d d \dots t o g e t \dots I = \frac{I_{1} - I_{2}}{2}