calculate-0-2-t-1-t-5-3-dt-

Question Number 35983 by abdo mathsup 649 cc last updated on 26/May/18

c a l c u l a t e \int_{0}^{\infty} \frac{2 \sqrt{t} + 1}{t^{5} + 3} d t .

Commented by prof Abdo imad last updated on 27/May/18

l e t p u t I = \int_{0}^{\infty} \frac{2 \sqrt{t} + 1}{t^{5} + 3} d t c h a n g e m e n t \sqrt{t} = x

g i v e I = \int_{0}^{\infty} \frac{2 x + 1}{x^{10} + 3} 2 x d x = 4 \int_{0}^{\infty} \frac{x^{2}}{x^{10} + 3} d x

+ 2 \int_{0}^{\infty} \frac{x}{x^{10} + 3} d x b u t c h a n g e m e n t x = 3^{\frac{1}{10}} u g i v e

\int_{0}^{\infty} \frac{x^{2}}{x^{10} + 3} d x = \int_{0}^{\infty} \frac{3^{\frac{1}{5}} u^{2}}{3 u^{10} + 3} 3^{\frac{1}{10}} d u

= \frac{3^{\frac{3}{10}}}{3} \int_{0}^{\infty} \frac{u^{2}}{1 + u^{10}} d u t h e n w e u s e t h e c h a n g .

u^{10} = t \Rightarrow u = t^{\frac{1}{10}} \Rightarrow \int_{0}^{\infty} \frac{u^{2}}{1 + u^{10}} d u

= \int_{0}^{\infty} \frac{t^{\frac{1}{5}}}{1 + t} \frac{1}{10} t^{\frac{1}{10} - 1} d t = \frac{1}{10} \int_{0}^{\infty} \frac{t^{\frac{3}{10} - 1}}{1 + t} d t

= \frac{1}{10} \frac{π}{s i n (\frac{3 π}{10})} b y t h e s a m e c h a n g . w e g e t

\int_{0}^{\infty} \frac{x}{x^{10} + 3} d x = \int_{0}^{\infty} \frac{3^{\frac{1}{10}} u}{3 u^{10} + 3} 3^{\frac{1}{10}} d u

= \frac{3^{\frac{1}{5}}}{3} \int_{0}^{\infty} \frac{u d u}{1 + u^{10}} = 3^{\frac{- 4}{5}} \int_{0}^{\infty} \frac{t^{\frac{1}{10}}}{1 + t} \frac{1}{10} t^{\frac{1}{10} - 1} d t

= \frac{3^{- \frac{4}{5}}}{10} \int_{0}^{\infty} \frac{t^{\frac{2}{5} - 1}}{1 + t} d t = \frac{3^{- \frac{4}{5}}}{10} \frac{π}{s i n (\frac{2 π}{5})} \Rightarrow

I = \frac{4}{10} \frac{π}{s i n (\frac{3 π}{10})} . \frac{3^{\frac{3}{10}}}{3} + 2 \frac{3^{- \frac{4}{5}}}{10} \frac{π}{s i n (\frac{2 π}{5})}

I = \frac{2 π}{5} 3^{- \frac{7}{10}} \frac{1}{s i n (\frac{3 π}{10})} + \frac{π}{5} 3^{- \frac{4}{5}} . \frac{1}{s i n (\frac{2 π}{5})}

I h a v e u s e d t h e f o r m u l a

\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 .