This is of the form ∫_(−𝑎)^𝑎▒𝑓(𝑥)𝑑𝑥 where 𝒇(𝒙)=𝐥𝐨𝐠((𝟐−𝒙)/(𝟐+𝒙)) And, 𝒇(−𝒙)=log((2 − (−𝑥))/(2 + (−𝑥))) 𝑓(−𝑥)=log((2 + 𝑥)/(2 − 𝑥)) Using log(𝑎/𝑏) = log a – log b 𝑓(−𝑥)=log⁡〖(2+𝑥)−log⁡(2−𝑥)〗 𝑓(−𝑥)=〖−(log〗⁡〖( 2−𝑥)−log⁡(2+𝑥)〗) 𝒇(−𝒙)=−𝐥𝐨𝐠((𝟐 − 𝒙)/(𝟐 + 𝒙)) Thus, 𝑓(−𝑥)=−𝑓(𝑥) ∴∫_(−1)^1 log((2 − 𝑥)/(2 + 𝑥))𝑑𝑥=0

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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