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

Question Number 66326 by mathmax by abdo last updated on 12/Aug/19

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

Commented by Prithwish sen last updated on 13/Aug/19

\frac{x^{4} + 1}{x^{6} + 1} = \frac{A}{x^{2} + 1} + \frac{B}{x^{2} + \sqrt{3} x + 1} + \frac{C}{x^{2} - \sqrt{3} x + 1}

x^{4} + 1 = x^{4} (A + B + C) + x^{3} (\sqrt{3} C - \sqrt{3} B) + x^{2} (2 B + 2 C - A) + x (\sqrt{3} C - \sqrt{3} B) + (A + B + C)

\Rightarrow A + B + C = 1

\sqrt{3} C - \sqrt{3} B = 0

2 B + 2 C - A = 0

A + B + C = 0

A = \frac{2}{3} B = \frac{1}{6} C = \frac{1}{6}

\int \frac{2 dx}{3 (1 + x^{2})} + \frac{1}{6} \int \frac{dx}{{(x + \frac{\sqrt{3}}{2})}^{2} + {(\frac{1}{2})}^{2}} + \frac{1}{6} \int \frac{dx}{{(x - \frac{\sqrt{3}}{2})}^{2} + {(\frac{1}{2})}^{2}}

Commented by mathmax by abdo last updated on 13/Aug/19

t h a n k y o u s i r .

Commented by Prithwish sen last updated on 13/Aug/19

welome

Commented by mathmax by abdo last updated on 13/Aug/19

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

\int_{1}^{+ \infty} \frac{x^{4} + 1}{x^{6} + 1} d x =_{x = \frac{1}{t}} - \int_{0}^{1} \frac{\frac{1}{t^{4}} + 1}{\frac{1}{t^{6}} + 1} (- \frac{d t}{t^{2}}) = \int_{0}^{1} \frac{1 + t^{4}}{1 + t^{6}} d t \Rightarrow

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

c h a n g e m e n t x = t^{\frac{1}{6}} g i v e \int_{0}^{\infty} \frac{x^{4}}{1 + x^{6}} d x = \int_{0}^{\infty} \frac{t^{\frac{2}{3}}}{1 + t} \frac{1}{6} t^{\frac{1}{6} - 1} d t

= \frac{1}{6} \int_{0}^{\infty} \frac{t^{\frac{2}{3} + \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} \times 2 = \frac{π}{3}

\int_{0}^{\infty} \frac{d x}{1 + x^{6}} =_{x = t^{\frac{1}{6}}} \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})} = \frac{π}{6} \times 2 = \frac{π}{3} \Rightarrow

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