Evaluate: โˆซ_(-1)^1โ€Šlog((2-x)/(2+x))dx

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[Sample Paper] Evaluate: โˆซ log (2-x/2+x) dx from -2 to 2 - Teachoo - CBSE Class 12 Sample Paper for 2024 Boards

part 2 - Question 24 - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Transcript

This is of the form โˆซ_(โˆ’๐‘Ž)^๐‘Žโ–’๐‘“(๐‘ฅ)๐‘‘๐‘ฅ where ๐’‡(๐’™)=๐ฅ๐จ๐ ((๐Ÿโˆ’๐’™)/(๐Ÿ+๐’™)) And, ๐’‡(โˆ’๐’™)=log((2 โˆ’ (โˆ’๐‘ฅ))/(2 + (โˆ’๐‘ฅ))) ๐‘“(โˆ’๐‘ฅ)=log((2 + ๐‘ฅ)/(2 โˆ’ ๐‘ฅ)) Using log(๐‘Ž/๐‘) = log a โ€“ log b ๐‘“(โˆ’๐‘ฅ)=logโกใ€–(2+๐‘ฅ)โˆ’logโก(2โˆ’๐‘ฅ)ใ€— ๐‘“(โˆ’๐‘ฅ)=ใ€–โˆ’(logใ€—โกใ€–( 2โˆ’๐‘ฅ)โˆ’logโก(2+๐‘ฅ)ใ€—) ๐’‡(โˆ’๐’™)=โˆ’๐ฅ๐จ๐ ((๐Ÿ โˆ’ ๐’™)/(๐Ÿ + ๐’™)) Thus, ๐‘“(โˆ’๐‘ฅ)=โˆ’๐‘“(๐‘ฅ) โˆดโˆซ_(โˆ’1)^1โ€Šlog((2 โˆ’ ๐‘ฅ)/(2 + ๐‘ฅ))๐‘‘๐‘ฅ=0

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