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

Question Number 67310 by mathmax by abdo last updated on 25/Aug/19

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

Commented by mathmax by abdo last updated on 25/Aug/19

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

w e h a v e x^{4} + x^{2} + 1 = {(x^{2} + 1)}^{2} - 2 x^{2} + x^{2} = {(x^{2} + 1)}^{2} - x^{2}

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

l e t d e c o m p o s e F (x) = \frac{1}{(x^{2} + x + 1) (x^{2} - x + 1)}

F (x) = \frac{a x + b}{x^{2} + x + 1} + \frac{c x + d}{x^{2} - x + 1}

F (- x) = F (x) \Rightarrow \frac{- a x + b}{x^{2} - x + 1} + \frac{- c x + d}{x^{2} + x + 1} = F (x) \Rightarrow

c = - a a n d d = b \Rightarrow F (x) = \frac{a x + b}{x^{2} + x + 1} + \frac{- a x + b}{x^{2} - x + 1}

F (0) = 1 = 2 b \Rightarrow b = \frac{1}{2}

F (1) = \frac{1}{3} = \frac{a + b}{3} + b - a \Rightarrow 3 = a + b + 3 b - 3 a = - 2 a + 2 \Rightarrow 1 = - 2 a \Rightarrow

a = - \frac{1}{2} \Rightarrow F (x) = \frac{- \frac{1}{2} x + \frac{1}{2}}{x^{2} + x + 1} + \frac{\frac{1}{2} x + \frac{1}{2}}{x^{2} - x + 1}

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

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

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

= \frac{1}{4} {[l n ∣ \frac{x^{2} - x + 1}{x^{2} + x + 1} ∣]}_{0}^{+ \infty} + \frac{3}{4} \int_{0}^{\infty} \frac{d x}{x^{2} - x + 1} + \frac{3}{4} \int_{0}^{\infty} \frac{d x}{x^{2} + x + 1}

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

\int_{0}^{\infty} \frac{d x}{x^{2} - x + 1} = \int_{0}^{\infty} \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} =_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{- \frac{1}{\sqrt{3}}}^{+ \infty} \frac{1}{u^{2} + 1} \frac{\sqrt{3}}{2} d u

= \frac{2}{\sqrt{3}} {[a r c t a n u]}_{- \frac{1}{\sqrt{3}}}^{+ \infty} = \frac{2}{\sqrt{3}} {\frac{π}{2} + a r c t a n (\frac{1}{\sqrt{3}})} = \frac{2}{\sqrt{3}} {\frac{π}{2} + \frac{π}{6}}

= \frac{2}{\sqrt{3}} {\frac{2 π}{3}} = \frac{4 π}{3 \sqrt{3}}

\int_{0}^{\infty} \frac{d x}{x^{2} + x + 1} = \int_{0}^{\infty} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{\frac{1}{\sqrt{3}}}^{\infty} \frac{1}{u^{2} + 1} \frac{\sqrt{3}}{2} d u

= \frac{2 \sqrt{3}}{3} {[a r c t a n (u)]}_{\frac{1}{\sqrt{3}}}^{+ \infty} = \frac{2}{\sqrt{3}} {\frac{π}{2} - \frac{π}{6}} = \frac{2}{\sqrt{3}} {\frac{π}{3}} = \frac{2 π}{3 \sqrt{3}} \Rightarrow

\int_{0}^{\infty} F (x) d x = \frac{3}{4} \times \frac{4 π}{3 \sqrt{3}} + \frac{3}{4} \times \frac{2 π}{3 \sqrt{3}} = \frac{π}{\sqrt{3}} + \frac{π}{2 \sqrt{3}} = \frac{3}{2} \frac{π}{\sqrt{3}}

w e h a v e I = 2 \int_{0}^{\infty} F (x) d x = \frac{3 π}{\sqrt{3}} = π \sqrt{3}

Commented by MJS last updated on 26/Aug/19

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

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

= - \frac{1}{4} \ln (x^{2} - x + 1) + \frac{\sqrt{3}}{6} \arctan \frac{\sqrt{3} (2 x - 1)}{3} + \frac{1}{4} \ln (x^{2} + x + 1) + \frac{\sqrt{3}}{6} \arctan \frac{\sqrt{3} (2 x + 1)}{3} =

= \frac{1}{4} \ln \frac{x^{2} + x + 1}{x^{2} - x + 1} + \frac{\sqrt{3}}{6} (\arctan \frac{\sqrt{3} (2 x - 1)}{3} + \arctan \frac{\sqrt{3} (2 x + 1)}{3}) + C

\underset{- \infty}{\int^{+ \infty}} \frac{d x}{x^{4} + x^{2} + 1} = \frac{π \sqrt{3}}{3}

Answered by Kunal12588 last updated on 25/Aug/19

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

(x + \frac{1}{x}) = t \Rightarrow (1 - \frac{1}{x^{2}}) d x = d t

(x - \frac{1}{x}) = u \Rightarrow (1 + \frac{1}{x^{2}}) d x = d u

x^{2} + \frac{1}{x^{2}} + 1 = t^{2} - 1 a n d x^{2} + \frac{1}{x^{2}} + 1 = u^{2} + 3

I = \frac{1}{2} \int \frac{d u}{u^{2} + 3} - \frac{1}{2} \int \frac{d t}{t^{2} - 1}

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

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

I_{1} = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} + x^{2} + 1} = {[I]}_{- \infty}^{+ \infty}

\underset{x \to + \infty}{l i m} I = \frac{1}{2 \sqrt{3}} \underset{x \to + \infty}{l i m a r c t a n} (\frac{x}{\sqrt{3}} - \frac{1}{x \sqrt{3}}) - \frac{1}{4} \underset{x \to + \infty}{l i m} l n ∣ \frac{1 - \frac{1}{x} + \frac{1}{x^{2}}}{1 + \frac{1}{x} + \frac{1}{x^{2}}} ∣

= \frac{1}{2 \sqrt{3}} \underset{x \to + \infty}{l i m a r c t a n} (\frac{x}{\sqrt{3}} - \frac{1}{x \sqrt{3}}) - \frac{1}{4} \underset{x \to + \infty}{l i m} l n ∣ 1 ∣

= \frac{1}{2 \sqrt{3}} \times \frac{π}{2} - 0 = \frac{π}{4 \sqrt{3}}

\underset{x \to - \infty}{l i m I} = \frac{1}{2 \sqrt{3}} \underset{x \to - \infty}{l i m} a r c t a n (\frac{x}{\sqrt{3}} - \frac{1}{x \sqrt{3}}) - 0

= \frac{1}{2 \sqrt{3}} \times \frac{- π}{2} = - \frac{π}{4 \sqrt{3}}

I_{1} = \frac{π}{4 \sqrt{3}} + \frac{π}{4 \sqrt{3}} = \frac{π}{2 \sqrt{3}}

Commented by mathmax by abdo last updated on 25/Aug/19

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

Commented by Kunal12588 last updated on 26/Aug/19

b u t i t s e e m s w r o n g . c a n u p l s c h e c k w h e r e i t

g o e s w r o n g

Commented by mathmax by abdo last updated on 26/Aug/19

g i v e o p p o r t u n i t y t o t h e m e t h o d \dots