prove-that-0-1-x-2-6-x-2-4-x-2-9-20-

Question Number 43545 by peter frank last updated on 11/Sep/18

p r o v e t h a t \int \underset{0}{\overset{1}{}} \frac{x^{2} + 6}{(x^{2} + 4) (x^{2} + 9)} = \frac{Π}{20}

Answered by behi83417@gmail.com last updated on 12/Sep/18

I = \frac{1}{2} \int \frac{2 x^{2} + 12}{(x^{2} + 4) (x^{2} + 9)} d x =

= \frac{1}{2} [\int (\frac{1}{x^{2} + 9} + \frac{1}{x^{2} + 4} - \frac{1}{5} (\frac{1}{x^{2} + 4} - \frac{1}{x^{2} + 9})] d x =

= \frac{3}{5} \int \frac{d x}{x^{2} + 9} + \frac{2}{5} \int \frac{d x}{x^{2} + 4} =

= \frac{3}{5} . \frac{1}{3} {t g}^{- 1} (\frac{x}{3}) + \frac{2}{5} . \frac{1}{2} {t g}^{- 1} (\frac{x}{2}) + c =

= \frac{1}{5} {t g}^{- 1} \frac{\frac{x}{3} + \frac{x}{2}}{1 - \frac{x^{2}}{6}} = \frac{1}{5} {t g}^{- 1} \frac{5 x}{6 - x^{2}} + c .

I = F (1) - F (0) = \frac{1}{5} [{t g}^{- 1} \frac{5}{5} - 0] =

= \frac{1}{5} . \frac{π}{4} = \frac{π}{20} .