Question 20

Evaluate: Definite Integration -1 → 1  ∫ (x + |x| + 1) / (x + 2|x| + 1)

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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 20 Evaluate: โˆซ1_(โˆ’1)^1โ–’(๐‘ฅ+|๐‘ฅ|+1)/(๐‘ฅ^2+2|๐‘ฅ|+1) We know that |๐‘ฅ|={โ–ˆ(โˆ’&๐‘ฅ, ๐‘ฅ<0@&๐‘ฅ, ๐‘ฅโ‰ฅ0)โ”ค So, for โ€“1 to 0, |x| = โ€“x and for 0 to 1 |x| = x So, our integral becomes โˆซ1_(โˆ’1)^1โ–’(๐‘ฅ+|๐‘ฅ|+1)/(๐‘ฅ^2+2|๐‘ฅ|+1) โˆซ1_(โˆ’1)^0โ–’(๐‘ฅ+|๐‘ฅ|+1)/(๐‘ฅ^2+2|๐‘ฅ|+1) โˆซ1_0^1โ–’(๐‘ฅ+|๐‘ฅ|+1)/(๐‘ฅ^2+2|๐‘ฅ|+1) โˆซ1_(โˆ’1)^0โ–’(๐‘ฅโˆ’๐‘ฅ+1)/(๐‘ฅ^2โˆ’2๐‘ฅ+1) โˆซ1_0^1โ–’(๐‘ฅ+๐‘ฅ+1)/(๐‘ฅ^2+2๐‘ฅ+1) โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅ^2โˆ’2๐‘ฅ+1) โˆซ1_0^1โ–’(2๐‘ฅ+1)/(๐‘ฅ^2+2๐‘ฅ+1) โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_0^1โ–’(2๐‘ฅ+1)/(๐‘ฅ+1)^2 โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_0^1โ–’(2๐‘ฅ+1+1โˆ’1)/(๐‘ฅ+1)^2 โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_0^1โ–’(2๐‘ฅ+2โˆ’1)/(๐‘ฅ+1)^2 โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_0^1โ–’(2(๐‘ฅ+1)โˆ’1)/(๐‘ฅ+1)^2 โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_0^1โ–’(2(๐‘ฅ+1) )/(๐‘ฅ+1)^2 โˆซ1_0^1โ–’1/(๐‘ฅ+1)^2 โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 โˆซ1_0^1โ–’(2 )/((๐‘ฅ+1) ) โˆซ1_0^1โ–’1/(๐‘ฅ+1)^2 โˆซ1_(โˆ’1)^0โ–’1/(๐‘ฅโˆ’1)^2 2 โˆซ1_0^1โ–’"dx " /((๐‘ฅ+1) ) โˆซ1_0^1โ–’1/(๐‘ฅ+1)^2 [(๐‘ฅโˆ’1)^(โˆ’1)/(โˆ’1)]_(โˆ’1)^0 2[lnโกใ€–|๐‘ฅ+1|ใ€— ]_0^1 [(๐‘ฅ+1)^(โˆ’1)/(โˆ’1)]_0^1 [(โˆ’1)/((๐‘ฅโˆ’1))]_(โˆ’1)^0 "2" [lnโกใ€–|๐‘ฅ+1|ใ€— ]_0^1 [(โˆ’1)/((๐‘ฅ+1))]_0^1 [(โˆ’1)/((0โˆ’1))โˆ’(โˆ’1)/((โˆ’1โˆ’1))] 2[lnโกใ€–|2|ใ€—โˆ’logโกใ€–|1|ใ€— ] [(โˆ’1)/((1+1))โˆ’(โˆ’1)/((0+1))] [1โˆ’1/2] 2[lnโกใ€–|2|ใ€—โˆ’0] [โˆ’1/2+1] 1/2+2 lnโก2โˆ’1/2 2 lnโก2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.