calculate-0-1-x-4-1-x-6-dx-

Question Number 33026 by prof Abdo imad last updated on 09/Apr/18

c a l c u l a t e \int_{0}^{\infty} \frac{1 + x^{4}}{1 + x^{6}} d x .

Commented by abdo imad last updated on 09/Apr/18

l e t p u t I = \int_{0}^{\infty} \frac{1 + x^{4}}{1 + x^{6}} d x w e h a v e

I = \int_{0}^{\infty} \frac{d x}{1 + x^{6}} + \int_{0}^{\infty} \frac{x^{4}}{1 + x^{6}} d x . t h e c h . x^{6} = t g i v e

\int_{0}^{\infty} \frac{d x}{1 + x^{6}} = \int_{0}^{\infty} \frac{\frac{1}{6} t^{\frac{1}{6} - 1}}{1 + t} d t = \frac{1}{6} \int_{0}^{\infty} \frac{t^{\frac{1}{6} - 1}}{1 + t} d t

= \frac{1}{6} \frac{π}{s i n (\frac{π}{6})} (b y u s i n g t h e r e s u l t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)}

w i t h 0 < a < 1)

= \frac{π}{6 . \frac{1}{2}} = \frac{π}{3} . a l s o t h e s a m e c h . x^{6} = t g i v e x = t^{\frac{1}{6}}

\int_{0}^{\infty} \frac{x^{4}}{1 + x^{6}} d x = \int_{0}^{\infty} \frac{t^{\frac{4}{6}}}{1 + t} . \frac{1}{6} t^{\frac{1}{6} - 1} d t

= \frac{1}{6} \int_{0}^{\infty} \frac{t^{\frac{4}{6} + \frac{1}{6} - 1}}{1 + t} d t = \frac{1}{6} \int_{0}^{\infty} \frac{t^{\frac{5}{6} - 1}}{1 + t} d t

= \frac{1}{6} \frac{π}{s i n (\frac{5 π}{6})} = \frac{π}{6 s i n (π - \frac{π}{6})} = \frac{π}{6 s i n (\frac{π}{6})} = \frac{π}{3} \Rightarrow

I = \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3} .

Answered by Joel578 last updated on 09/Apr/18

I = \lim_{n \to \infty} (\int_{0}^{n} \frac{x^{4} + 1}{x^{6} + 1} d x)

\int \frac{x^{4} + 1}{x^{6} + 1} d x = \int \frac{x^{4} + 1 + x^{2} - x^{2}}{(x^{2} + 1) (x^{4} - x^{2} + 1)} d x

= \int \frac{1}{x^{2} + 1} d x + \int \frac{x^{2}}{x^{6} + 1} d x

= \tan^{- 1} x + \int \frac{x^{2}}{x^{6} + 1} d x (u = x^{3} \to d u = 3 x^{2} d x)

= \tan^{- 1} x + \frac{1}{3} \int \frac{1}{u^{2} + 1} d u

= \tan^{- 1} x + \frac{1}{3} \tan^{- 1} u + C

= \tan^{- 1} (x) + \frac{1}{3} \tan^{- 1} (x^{3}) + C

I = \lim_{n \to \infty} {[\tan^{- 1} (x) + \frac{1}{3} \tan^{- 1} (x^{3})]}_{0}^{n}

= \lim_{n \to \infty} (\tan^{- 1} (n) + \frac{1}{3} \tan^{- 1} (n^{3})) - (\tan^{- 1} (0) + \frac{1}{3} \tan^{- 1} (0))

= \frac{π}{2} + \frac{1}{3} (\frac{π}{2}) - 0

= \frac{2 π}{3}

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