Ex 7.5, 20 - Integrate 1/ x (x4 - 1) - Chapter 7 CBSE - Ex 7.5

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.5, 20 1/(๐‘ฅ(๐‘ฅ4โˆ’1) ) Multiplying integrand by ๐‘ฅ^3/๐‘ฅ^3 1/(๐‘ฅ(๐‘ฅ4 โˆ’ 1) ) = 1/(๐‘ฅ(๐‘ฅ^4 โˆ’ 1) ) ร— ๐‘ฅ^3/๐‘ฅ^3 = ๐‘ฅ^3/(๐‘ฅ^4 (๐‘ฅ^4 โˆ’ 1) ) Let t = ๐‘ฅ^4 Differentiating both sides ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ก/๐‘‘๐‘ฅ = 4๐‘ฅ^3 ๐‘‘๐‘ก/(4๐‘ฅ^3 ) = ๐‘‘๐‘ฅ Substituting value of ๐‘ก = ๐‘ฅ^4 & ๐‘‘๐‘ฅ ๐‘‘๐‘ก/(4๐‘ฅ^3 ) " " โˆซ1โ–’๐‘ฅ^3/(๐‘ฅ^4 (๐‘ฅ^4โˆ’ 1) ) ๐‘‘๐‘ฅ = โˆซ1โ–’๐‘ฅ^3/(๐‘ก(๐‘ก โˆ’ 1) ) ๐‘‘๐‘ก/(4๐‘ฅ^3 ) " " = 1/4 โˆซ1โ–’๐‘‘๐‘ก/(๐‘ก(๐‘ก โˆ’ 1) ) We can write integrand as 1/(๐‘ก(๐‘ก โˆ’ 1) ) = ๐ด/๐‘ก + ๐ต/(๐‘ก โˆ’ 1) 1/(๐‘ก(๐‘ก โˆ’ 1) ) = (๐ด(๐‘ก โˆ’ 1) + ๐ต ๐‘ก)/๐‘ก(๐‘ก โˆ’ 1) By cancelling denominator 1 = ๐ด(๐‘กโˆ’1)+๐ต๐‘ก Putting t = 0 in (1) 1 = ๐ด(0โˆ’1)+๐ตร—0 1 = ๐ดร—(โˆ’1) 1 = โˆ’๐ด ๐ด = โˆ’1 Similarly putting t = 1 in (1) 1 = A(tโˆ’1)+Bt 1 = ๐ด(1โˆ’1)+๐ตร—1 1 = ๐ดร—0+๐ต 1 = ๐ต ๐ต = 1 Therefore 1/4 โˆซ1โ–’1/(๐‘ก(๐‘ก โˆ’ 1) ) ๐‘‘๐‘ก = โˆซ1โ–’(โˆ’1)/(๐‘ก ) ๐‘‘๐‘ก + โˆซ1โ–’1/(๐‘ก โˆ’ 1 ) = โˆ’ใ€–log ใ€—โก|๐‘ก|+ใ€–log ใ€—โก|๐‘กโˆ’1|+๐ถ = ใ€–log ใ€—โก|(๐‘ก โˆ’ 1)/๐‘ก|+๐ถ Putting t =ใ€– ๐‘ฅใ€—^4 โˆซ1โ–’1/(๐‘ฅ(๐‘ฅ^4 โˆ’ 1) ) ๐‘‘๐‘ฅ = ๐Ÿ/๐Ÿ’ ใ€–๐ฅ๐จ๐  ใ€—โก|(๐’™^๐Ÿ’ โˆ’ ๐Ÿ)/๐’™^๐Ÿ’ |+๐‘ช

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