calculate-0-dx-x-4-x-2-1-

Question Number 93634 by abdomathmax last updated on 14/May/20

c a l c u l a t e \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1}

Commented by mathmax by abdo last updated on 14/May/20

I = \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + 1} l e t φ (z) = \frac{1}{z^{4} + z^{2} + 1}

p o l e s o f φ ?

z^{4} + z^{2} + 1 = 0 \to t^{2} + t + 1 = 0 (t = z^{2})

Δ = - 3 \Rightarrow t_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}}, t_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}} \Rightarrow

φ (z) = \frac{1}{(z^{2} - e^{\frac{i 2 π}{3}}) (z^{2} - e^{- \frac{i 2 π}{3}})} = \frac{1}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{\frac{- i π}{3}}) (z + e^{- \frac{i π}{3}})}

r d s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{3}}) + R e s (φ, - e^{- \frac{i π}{3}})}

R e s (φ, e^{\frac{i π}{3}}) = \frac{1}{2 e^{\frac{i π}{3}} (2 i s i n (\frac{2 π}{3}))} = \frac{e^{- \frac{i π}{3}}}{4 i \times \frac{\sqrt{3}}{2}} = \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}

R e s (φ, - e^{- \frac{i π}{3}}) = \frac{1}{- 2 e^{- \frac{i π}{3}} (- 2 i s i n (\frac{2 π}{3}))} = \frac{e^{\frac{i π}{3}}}{4 i \times \frac{\sqrt{3}}{2}} = \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{2 i \sqrt{3}} {e^{- \frac{i π}{3}} + e^{\frac{i π}{3}}} = \frac{π}{\sqrt{3}} (2 c o s (\frac{π}{3})) = \frac{π}{\sqrt{3}}

2 \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{\sqrt{3}} \Rightarrow ★ \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{2 \sqrt{3}} ★

Commented by mathmax by abdo last updated on 14/May/20

s o r r y 2 \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{\sqrt{3}} \Rightarrow ★ \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{2 \sqrt{3}} ★

Answered by Kunal12588 last updated on 14/May/20

(x^{4} + x^{2} + 1) = ({a x}^{2} + b x + c) ({p x}^{2} + q x + r)

x^{4} + x^{2} + 1 = {a p x}^{4} + (a q + b p) x^{3} + (a r + b q + c p) x^{2}

+ (b r + c q) x + c r

a = 1, b = \pm 1, c = 1

p = 1, q = \mp 1, r = 1

x^{4} + x^{2} + 1 = (x^{2} + x + 1) (x^{2} - x + 1)

\frac{1}{x^{4} + x^{2} + 1} = \frac{a x + b}{x^{2} + x + 1} + \frac{p x + q}{x^{2} - x + 1}

1 = (a + p) x^{3} + (- a + b + p + q) x^{2} + (a - b + p + q) x

+ (b + q)

a = \frac{1}{2}, b = \frac{1}{2}

p = \frac{- 1}{2}, q = \frac{1}{2}

\frac{1}{x^{4} + x^{2} + 1} = \frac{x + 1}{x^{2} + x + 1} - \frac{x - 1}{x^{4} + x + 1}

I = \int \frac{d x}{x^{4} + x^{2} + 1} = \frac{1}{2} \int \frac{x + 1}{x^{2} + x + 1} d x - \frac{1}{2} \int \frac{x - 1}{x^{2} + x + 1} d x

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

= \frac{1}{4} \int \frac{2 x + 1}{x^{2} + x + 1} d x + \frac{1}{4} \int \frac{d x}{x^{2} + x + 1}

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

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

I_{1} = \frac{1}{4} l n ∣ x^{2} + x + 1 ∣ + \frac{1}{2 \sqrt{3}} {t a n}^{- 1} \frac{2 x + 1}{\sqrt{3}} + m

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

= \frac{1}{4} \int \frac{2 x - 1}{x^{2} - x + 1} d x - \frac{1}{4} \int \frac{d x}{x^{2} - x + 1}

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

I_{2} = \frac{1}{4} l n ∣ x^{2} - x + 1 ∣ - \frac{1}{2 \sqrt{3}} {t a n}^{- 1} \frac{2 x - 1}{\sqrt{3}} + n

I = \frac{1}{4} l n ∣ x^{2} + x + 1 ∣ + \frac{1}{2 \sqrt{3}} {t a n}^{- 1} \frac{2 x + 1}{\sqrt{3}}

- \frac{1}{4} l n ∣ x^{2} - x + 1 ∣ + \frac{1}{2 \sqrt{3}} {t a n}^{- 1} \frac{2 x - 1}{\sqrt{3}} + C

= \frac{1}{4} l n ∣ \frac{x^{2} + x + 1}{x^{2} - x + 1} ∣ + \frac{1}{2 \sqrt{3}} [{t a n}^{- 1} \frac{2 x + 1}{\sqrt{3}} + {t a n}^{- 1} \frac{2 x - 1}{\sqrt{3}}] + C

\int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{1}{4} {[l n ∣ \frac{x^{2} + x + 1}{x^{2} - x + 1} ∣]}_{0}^{\infty} +

+ \frac{1}{2 \sqrt{3}} {[{t a n}^{- 1} \frac{2 x + 1}{\sqrt{3}} + {t a n}^{- 1} \frac{2 x - 1}{\sqrt{3}}]}_{0}^{\infty}

= \frac{1}{4} {l n (\frac{1 + \frac{1}{\infty} + \frac{1}{\infty^{2}}}{1 - \frac{1}{\infty} + \frac{1}{\infty^{2}}}) - l n (\frac{0^{2} + 0 + 1}{0^{2} - 0 + 1})}

+ \frac{1}{2 \sqrt{3}} {{t a n}^{- 1} (\infty) + {t a n}^{- 1} (\infty) - {t a n}^{- 1} (\frac{1}{\sqrt{3}}) - {t a n}^{- 1} (\frac{- 1}{\sqrt{3}})}

= \frac{1}{4} [l n (1) - l n (1)] + \frac{1}{2 \sqrt{3}} [\frac{π}{2} + \frac{π}{2} - \frac{π}{6} + \frac{π}{6}]

= \frac{π}{2 \sqrt{3}}

\int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π}{2 \sqrt{3}}

Commented by niroj last updated on 14/May/20

Great hardwork��

Commented by prakash jain last updated on 14/May/20

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Commented by Kunal12588 last updated on 14/May/20

thanks sir!

Commented by mathmax by abdo last updated on 14/May/20

t h a n k x f o r t h i s h a r d w o r k

Answered by Ar Brandon last updated on 14/May/20

L = \int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{1}{2} \int_{0}^{\infty} \frac{(x^{2} + 1) - (x^{2} - 1)}{x^{4} + x^{2} + 1} d x

\Rightarrow 2 L = \int_{0}^{\infty} \frac{x^{2} + 1}{x^{4} + x^{2} + 1} d x - \int_{0}^{\infty} \frac{x^{2} - 1}{x^{4} + x^{2} + 1} d x

\Rightarrow 2 L = \int_{0}^{\infty} \frac{1 + \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} d x - \int_{0}^{\infty} \frac{1 - \frac{1}{x^{2}}}{x^{2} + 1 + \frac{1}{x^{2}}} d x

\Rightarrow 2 L = \int_{0}^{\infty} \frac{1 + \frac{1}{x^{2}}}{{(x - \frac{1}{x})}^{2} + 3} d x - \int_{0}^{\infty} \frac{1 - \frac{1}{x^{2}}}{{(x + \frac{1}{x})}^{2} - 1} d x

\Rightarrow 2 L \underset{u = x - \frac{1}{x}}{\overset{v = x + \frac{1}{x}}{=}} \int_{- \infty}^{+ \infty} \frac{du}{u^{2} + 3} - \int_{+ \infty}^{+ \infty} \frac{dv}{v^{2} - 1}

2 L = \frac{\sqrt{3}}{3} {[\arctan (\frac{u}{\sqrt{3}})]}_{- \infty}^{+ \infty}

2 L = \frac{π \sqrt{3}}{3} \Rightarrow L = \frac{π \sqrt{3}}{6}

\int_{0}^{\infty} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π \sqrt{3}}{6}