calculate-x-4-x-8-16-dx-

Question Number 42695 by prof Abdo imad last updated on 31/Aug/18

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{x^{4}}{x^{8} + 16} d x

Commented by maxmathsup by imad last updated on 01/Sep/18

l e t A = \int_{- \infty}^{+ \infty} \frac{x^{4}}{x^{8} + 16} d x \Rightarrow A = 2 \int_{0}^{\infty} \frac{x^{4}}{x^{8} + {(\sqrt{2})}^{8}} c h a n g e m e n t x = t \sqrt{2} g i v e

A = 2 \int_{0}^{\infty} \frac{4 t^{4}}{16 t^{8} + 16} d t = \frac{1}{2} \int_{0}^{\infty} \frac{t^{4}}{1 + t^{8}} d t

=_{t = u^{\frac{1}{8}}} \frac{1}{2} \int_{0}^{\infty} \frac{u^{\frac{1}{2}}}{1 + u} \frac{1}{8} u^{\frac{1}{8} - 1} d u = \frac{1}{16} \int_{0}^{\infty} \frac{u^{\frac{1}{2} + \frac{1}{8} - 1}}{1 + u} d u

= \frac{1}{16} \int_{0}^{\infty} \frac{u^{\frac{5}{8} - 1}}{1 + u} d u = \frac{1}{16} \frac{π}{s i n (\frac{5 π}{8})} b u t s i n (\frac{5 π}{8}) = s i n (\frac{π}{2} + \frac{π}{8}) = c o s (\frac{π}{8}) = \frac{\sqrt{2 + \sqrt{2}}}{2}

\Rightarrow A = \frac{π}{16} \frac{\sqrt{2 + \sqrt{2}}}{2} \Rightarrow A = \frac{π \sqrt{2 + \sqrt{2}}}{32} .