Ex 7.5, 14 i.jpg

Ex 7.5, 14 ii.jpg
Ex 7.5, 14 iii.jpg

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

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 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.