Ex 7.5, 2 - Integrate 1 / x2 - 9 - Chapter 7 Class 12 NCERT - Ex 7.5

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

Transcript

Ex 7.5, 2 1/(๐‘ฅ2โˆ’ 9) Solving integrand 1/(๐‘ฅ2โˆ’ 9)=1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) ) We can write it as 1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) )=๐ด/((๐‘ฅ โˆ’ 3) ) + ๐ต/((๐‘ฅ + 3) ) 1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) )=(๐ด(๐‘ฅ + 3) + ๐ต(๐‘ฅ โˆ’ 3))/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) ) " " Cancelling denominator 1 = ๐ด(๐‘ฅ + 3) + ๐ต(๐‘ฅ โˆ’ 3) Putting ๐‘ฅ = 3 in (1) 1 = ๐ด(๐‘ฅ+3)+๐ต(๐‘ฅโˆ’3) 1 = ๐ด(3+3) + ๐ต(3โˆ’3) 1 = ๐ดร—6+ ๐ตร—0 1 = 6๐ด ๐ด = 1/6 Similarly Putting y=โˆ’3 in (1) 1 = ๐ด(โˆ’3+3) + ๐ต(โˆ’3โˆ’3) 1 = ๐ดร—0+ ๐ตร—(โˆ’6) 1 = โˆ’6๐ต ๐ต = (โˆ’1)/6 Hence we can write it as 1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) ) = 1/6(๐‘ฅ โˆ’ 3) โˆ’ 1/6(๐‘ฅ + 3) Therefore โˆซ1โ–’1/((๐‘ฅ โˆ’ 3) (๐‘ฅ + 3) ) ๐‘‘๐‘ฅ = 1/6 โˆซ1โ–’1/((๐‘ฅ โˆ’ 3) ) ๐‘‘๐‘ฅ โˆ’ 1/6 โˆซ1โ–’1/((๐‘ฅ + 3) ) ๐‘‘๐‘ฅ = 1/6 ใ€–log ใ€—โก|๐‘ฅโˆ’3|โˆ’ 1/6 ใ€–log ใ€—โก|๐‘ฅ+3|+๐ถ = 1/6 (ใ€–log ใ€—โก|๐‘ฅโˆ’3|โˆ’ใ€–log ใ€—โก|๐‘ฅ+3| )+๐ถ = ๐Ÿ/๐Ÿ” ใ€–๐’๐’๐’ˆ ใ€—โก|(๐’™ โˆ’ ๐Ÿ‘)/(๐’™ + ๐Ÿ‘)|+๐‘ช

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.