find-the-value-of-0-1-ln-1-1-x-2-2-x-2-dx-

Question Number 197821 by mnjuly1970 last updated on 30/Sep/23

f i n d t h e v a l u e o f :

Ω = \int_{0}^{1} \frac{\ln (1 + \frac{1}{x^{2}})}{2 + x^{2}} d x = ?

Answered by witcher3 last updated on 05/Oct/23

x \to \frac{1}{x}

Ω = \int_{1}^{\infty} \frac{\ln (1 + x^{2})}{({2 x}^{2} + 1)} dx

Ω (a) = \int_{1}^{\infty} \frac{\ln (1 + {ax}^{2})}{({2 x}^{2} + 1)}, Ω (a) = 0

Ω^{'} (a) = \int_{1}^{\infty} \frac{x^{2}}{(1 + {ax}^{2}) ({2 x}^{2} + 1)}

= \int_{1}^{\infty} \frac{{2 x}^{2} + 1 - (1 + {ax}^{2})}{(2 - a) ({2 x}^{2} + 1) (1 + {ax}^{2})}

= \frac{1}{2 - a} {\int_{1}^{\infty} \frac{dx}{1 + {ax}^{2}} - \int_{1}^{\infty} \frac{dx}{1 + {2 x}^{2}}}

= \frac{1}{(2 - a) \sqrt{a}} (\frac{π}{2} - \tan^{- 1} (\sqrt{a})) - \frac{1}{(2 - a) \sqrt{2}} (\frac{π}{2} - \tan^{- 1} (\sqrt{2}))

- \int_{0}^{1} \frac{\tan^{- 1} (\sqrt{a})}{\sqrt{a} (2 - a)} da

= - 2 \int_{0}^{1} \frac{\tan^{- 1} (a)}{2 - a^{2}} . . can solved {Li}_{2} . .

but non nice close formez

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