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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise


Misc 1 Integrate the function 1/(๐‘ฅ โˆ’ ๐‘ฅ^3 ) We can it write as 1/(๐‘ฅ โˆ’ ๐‘ฅ^3 )=(1/๐‘ฅ^3 )/(((๐‘ฅ โˆ’ ๐‘ฅ^3 ))/๐‘ฅ^3 ) =1/(๐‘ฅ^3 (๐‘ฅ/๐‘ฅ^3 โˆ’ ๐‘ฅ^3/๐‘ฅ^3 ) ) =1/(๐‘ฅ^3 (1/๐‘ฅ^2 โˆ’1) ) Dividing by ๐‘ฅ^3 both numerator and denominator Integrating w.r.t. ๐‘ฅ โˆซ1โ–’ใ€–1/(๐‘ฅ โˆ’ ๐‘ฅ^3 ) ๐‘‘๐‘ฅใ€—=โˆซ1โ–’1/(๐‘ฅ^3 (1/๐‘ฅ^2 โˆ’ 1) ) ๐‘‘๐‘ฅ Let t = 1/๐‘ฅ^2 โˆ’1 Differentiating with respect to ๐‘ฅ ๐‘‘/๐‘‘๐‘ฅ (1/๐‘ฅ^2 โˆ’1)=๐‘‘๐‘ก/๐‘‘๐‘ฅ (โˆ’2)/๐‘ฅ^3 =๐‘‘๐‘ก/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ=(๐‘ฅ^3 ๐‘‘๐‘ก)/(โˆ’2) Putting the value t and dt in the equation โˆซ1โ–’ใ€–1/(๐‘ฅ^3 (1/๐‘ฅ^2 โˆ’1) ) ๐‘‘๐‘ฅใ€—=โˆซ1โ–’ใ€–1/(๐‘ฅ^3 (๐‘ก) ) ร— (๐‘ฅ^3 ๐‘‘๐‘ก)/(โˆ’2)ใ€— =โˆซ1โ–’ใ€–1/(โˆ’2) ๐‘‘๐‘ก/๐‘กใ€— =(โˆ’1)/( 2) โˆซ1โ–’๐‘‘๐‘ก/๐‘ก =(โˆ’1)/( 2) ๐‘™๐‘œ๐‘”|๐‘ก|+๐ถ Putting back ๐‘ก=1/๐‘ฅ^2 โˆ’1 =(โˆ’1)/( 2) ๐‘™๐‘œ๐‘”|1/๐‘ฅ^2 โˆ’1|+๐ถ =(โˆ’1)/( 2) ๐‘™๐‘œ๐‘”|1/๐‘ฅ^2 โˆ’๐‘ฅ^2/๐‘ฅ^2 |+๐ถ =(โˆ’1)/( 2) ๐‘™๐‘œ๐‘”|(1 โˆ’ ๐‘ฅ^2)/๐‘ฅ^2 |+๐ถ = 1/2 log |(1 โˆ’ ๐‘ฅ^2)/๐‘ฅ^2 |^(โˆ’1)+๐ถ = ๐Ÿ/๐Ÿ log |๐’™^๐Ÿ/(๐Ÿ โˆ’ ๐’™^๐Ÿ )|+๐‘ช We know that a log b = log ๐‘^๐‘Ž

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.