Ex 7.5, 6.jpg

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.5, 6 (1 โˆ’ ๐‘ฅ2)/(๐‘ฅ(1 โˆ’ 2๐‘ฅ)) (1 โˆ’ ๐‘ฅ2)/(๐‘ฅ(1 โˆ’ 2๐‘ฅ)) = (1 โˆ’ ๐‘ฅ2)/(๐‘ฅ โˆ’ 2๐‘ฅ^2 ) Integrating โˆซ1โ–’(1 โˆ’ ๐‘ฅ^2)/(๐‘ฅ โˆ’ 2๐‘ฅ^2 ) ๐‘‘๐‘ฅ=โˆซ1โ–’(1/2 + (โˆ’ ๐‘ฅ/2 + 1)/(โˆ’2๐‘ฅ^2+ ๐‘ฅ)) ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’๐‘‘๐‘ฅ+1/2 โˆซ1โ–’(โˆ’๐‘ฅ + 2)/(โˆ’2๐‘ฅ^2+ ๐‘ฅ) ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’๐‘‘๐‘ฅโˆ’1/2 โˆซ1โ–’(๐‘ฅ โˆ’ 2)/๐‘ฅ(1 โˆ’ 2๐‘ฅ) ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’๐‘‘๐‘ฅโˆ’1/2 โˆซ1โ–’(๐‘ฅ โˆ’ 2)/๐‘ฅ(2๐‘ฅ โˆ’ 1) ๐‘‘๐‘ฅ Now Solving (๐‘ฅ โˆ’ 2)/(๐‘ฅ (2๐‘ฅ โˆ’ 1) ) = ๐ด/๐‘ฅ + ๐ต/(2๐‘ฅ โˆ’ 1) (๐‘ฅ โˆ’ 2)/(๐‘ฅ (2๐‘ฅ โˆ’ 1) ) = (๐ด(2๐‘ฅ โˆ’ 1) + ๐ต๐‘ฅ)/(๐‘ฅ (2๐‘ฅ โˆ’ 1) ) Cancelling denominator ๐‘ฅโˆ’2=๐ด(2๐‘ฅโˆ’1)+๐ต๐‘ฅ Putting x = 0, in (2) ๐‘ฅโˆ’2=๐ด(2๐‘ฅโˆ’1)+๐ต๐‘ฅ 0โˆ’2 = ๐ด(2ร—0โˆ’1) + ๐ตร—0 โˆ’2 = A(โˆ’1) โˆ’2 = โˆ’๐ด ๐ด = 2 Similarly Putting x = 1/2 , in (2) 1/2 โˆ’ 2 = A(1/2ร—2โˆ’1)+๐ตร—1/2 โˆ’ 3/2 = A(1โˆ’1)+๐ตร—1/2 โˆ’ 3/2 = Aร—0+ ๐ต/2 โˆ’ 3/2 = ๐ต/2 ๐ต = โˆ’3 Hence we can write it as (๐‘ฅ โˆ’ 2)/(๐‘ฅ (2๐‘ฅ โˆ’ 1) ) = 2/๐‘ฅ + ((โˆ’3))/(2๐‘ฅ โˆ’ 1) = 2/๐‘ฅ โˆ’ 3/(2๐‘ฅ โˆ’ 1) Therefore , from (1) we get, โˆซ1โ–’(1 โˆ’ ๐‘ฅ^2)/๐‘ฅ(1 โˆ’ 2๐‘ฅ) =1/2 โˆซ1โ–’๐‘‘๐‘ฅ+ 1/2 โˆซ1โ–’(2/๐‘ฅ โˆ’ 3/(2๐‘ฅ โˆ’ 1)) ๐‘‘๐‘ฅ =1/2 โˆซ1โ–’๐‘‘๐‘ฅ+ 2/2 โˆซ1โ–’ใ€–๐‘‘๐‘ฅ/๐‘ฅ โˆ’3/2ใ€— โˆซ1โ–’๐‘‘๐‘ฅ/(2๐‘ฅ โˆ’ 1) =1/2 โˆซ1โ–’๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–๐‘‘๐‘ฅ/๐‘ฅ + 3/2ใ€— โˆซ1โ–’๐‘‘๐‘ฅ/(1 โˆ’ 2๐‘ฅ) =1/2 ๐‘ฅ+logโก|๐‘ฅ|+3/2 logโก|1 โˆ’ 2๐‘ฅ|/(โˆ’2) +๐ถ =๐’™/๐Ÿ +๐’๐’๐’ˆโก|๐’™|โˆ’๐Ÿ‘/๐Ÿ’ ๐’๐’๐’ˆโก|๐Ÿโˆ’๐Ÿ๐’™|+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.