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Ex 7.5, 14 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 14 - Chapter 7 Class 12 Integrals - Part 3

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Transcript

Ex 7.5, 14 (3๐‘ฅ โˆ’ 1)/(๐‘ฅ + 2)^2 We can write it as (3๐‘ฅ โˆ’ 1)/(๐‘ฅ + 2)^2 = ๐ด/((๐‘ฅ + 2) )+๐ต/(๐‘ฅ + 2)^2 (3๐‘ฅ โˆ’ 1)/(๐‘ฅ + 2)^2 = (๐ด(๐‘ฅ + 2) + ๐ต)/(๐‘ฅ + 2)^2 By cancelling denominator 3๐‘ฅโˆ’1 = ๐ด(๐‘ฅ + 2)+๐ต Putting x = โˆ’2 in (1) 3๐‘ฅโˆ’1=๐ด (๐‘ฅ+2)+๐ต 3(โˆ’2)โˆ’1=๐ด (โˆ’2+2)+๐ต โˆ’6โˆ’1=๐ดร—0+๐ต โˆ’7=๐ต ๐ต=โˆ’7 Putting x = 0 in (1) 3๐‘ฅโˆ’1=๐ด (๐‘ฅ+2)+๐ต 3ร—0โˆ’1=๐ด (0+2)+๐ต โˆ’1=2๐ด+๐ต โˆ’1=2๐ด+(โˆ’7) โˆ’1+7=2๐ด 2๐ด=6 ๐ด=3 Hence we can write it as (3๐‘ฅ โˆ’ 1)/(๐‘ฅ + 2)^2 = 3/( ๐‘ฅ + 2)+(โˆ’7)/(๐‘ฅ + 2)^2 Integrating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ. โˆซ1โ–’(3๐‘ฅ โˆ’ 1)/(๐‘ฅ + 2)^2 ๐‘‘๐‘ฅ=โˆซ1โ–’(3/(๐‘ฅ + 2) โˆ’ 5/(๐‘ฅ + 2)^2 ) ๐‘‘๐‘ฅ =3โˆซ1โ–’ใ€–1/(๐‘ฅ + 2) ใ€— ๐‘‘๐‘ฅโˆ’7โˆซ1โ–’ใ€–1/(๐‘ฅ + 2)^2 ใ€— ๐‘‘๐‘ฅ =3 ใ€–log ใ€—โก|๐‘ฅ+2|โˆ’7โˆซ1โ–’ใ€–(๐‘ฅ+2)^(โˆ’2) ใ€— ๐‘‘๐‘ฅ =3 ใ€–log ใ€—โก|๐‘ฅ+2|โˆ’7 (๐‘ฅ + 2)^(โˆ’2 + 1)/(โˆ’2 + 1) =3 ใ€–log ใ€—โก|๐‘ฅ+2|โˆ’7 (๐‘ฅ + 2)^(โˆ’1)/(โˆ’1) +๐ถ =๐Ÿ‘ ใ€–๐ฅ๐จ๐  ใ€—โก|๐’™+๐Ÿ|+๐Ÿ•/((๐’™ + ๐Ÿ) ) +๐‘ช

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.