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

$$provethat\int \underset{0}{\stackrel{1}{}}\frac{x^2+6}{(x^2+4)(x^2+9)}=\frac{\mathrm{\Pi }}{20}$$

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

$$I=\frac{1}{2}\int \frac{2x^2+12}{(x^2+4)(x^2+9)}dx=$$$$=\frac{1}{2}[\int (\frac{1}{x^2+9}+\frac{1}{x^2+4}-\frac{1}{5}(\frac{1}{x^2+4}-\frac{1}{x^2+9})]dx=$$$$=\frac{3}{5}\int \frac{dx}{x^2+9}+\frac{2}{5}\int \frac{dx}{x^2+4}=$$$$=\frac{3}{5}.\frac{1}{3}{tg}^{-1}\left(\frac{x}{3}\right)+\frac{2}{5}.\frac{1}{2}{tg}^{-1}\left(\frac{x}{2}\right)+c=$$$$=\frac{1}{5}{tg}^{-1}\frac{\frac{x}{3}+\frac{x}{2}}{1-\frac{x^2}{6}}=\frac{1}{5}{tg}^{-1}\frac{5x}{6-x^2}+c.$$$$I=F\left(1\right)-F\left(0\right)=\frac{1}{5}[{tg}^{-1}\frac{5}{5}-0]=$$$$=\frac{1}{5}.\frac{\pi }{4}=\frac{\pi }{20}.◼$$

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