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Ex 7.5, 1 - Integrate x / (x + 1) (x + 2) - Chapter 7 - Integration by partial fraction - Type 1

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.5, 1 ๐‘ฅ/((๐‘ฅ + 1) (๐‘ฅ + 2) ) Solving integrand ๐‘ฅ/((๐‘ฅ + 1) (๐‘ฅ + 2) ) We can write it as ๐‘ฅ/((๐‘ฅ + 1) (๐‘ฅ + 2) ) " " = ๐ด/((๐‘ฅ + 1) ) + ๐ต/((๐‘ฅ + 2) ) ๐‘ฅ/((๐‘ฅ + 1) (๐‘ฅ + 2) ) " " = (๐ด(๐‘ฅ + 2) + ๐ต (๐‘ฅ + 1))/((๐‘ฅ + 1) (๐‘ฅ + 2) ) Cancelling denominator ๐‘ฅ = ๐ด(๐‘ฅ+2) + ๐ต(๐‘ฅ+1) Putting x=โˆ’1 in (1) โˆ’1=๐ด(โˆ’1+2) + ๐ต(โˆ’1+1) โˆ’1=๐ดร—1+ ๐ตร—0 โˆ’1=๐ด ๐ด=โˆ’1 Similarly Putting y=โˆ’2 in (1) โˆ’2 = ๐ด(โˆ’2+2) + ๐ต(โˆ’2+1) โˆ’2 = ๐ดร—0+ ๐ตร—(โˆ’1) โˆ’2 = โˆ’ ๐ต ๐ต = 2 Hence we can write it as ๐‘ฅ/((๐‘ฅ + 1) (๐‘ฅ + 2) ) = (โˆ’1)/((๐‘ฅ + 1) ) + 2/((๐‘ฅ + 2) ) Therefore โˆซ1โ–’๐‘ฅ/((๐‘ฅ + 1) (๐‘ฅ + 2) ) ๐‘‘๐‘ฅ = โˆซ1โ–’(โˆ’1)/((๐‘ฅ + 1) ) ๐‘‘๐‘ฅ + โˆซ1โ–’2/((๐‘ฅ + 2) ) ๐‘‘๐‘ฅ = โˆ’1โˆซ1โ–’1/((๐‘ฅ + 1) ) ๐‘‘๐‘ฅ + 2โˆซ1โ–’1/((๐‘ฅ + 2) ) ๐‘‘๐‘ฅ = โˆ’ใ€–log ใ€—โก|๐‘ฅ+1|+2 ใ€–log ใ€—โก|๐‘ฅ+2|+๐ถ = โˆ’ใ€–log ใ€—โก|๐‘ฅ+1|+ใ€–log ใ€—โกใ€–|๐‘ฅ+2|^2 ใ€—+๐ถ = ใ€–log ใ€—โก|(๐‘ฅ + 2)^2/(๐‘ฅ + 1)|+๐ถ = ใ€–๐ฅ๐จ๐  ใ€—โกใ€–(๐’™ + ๐Ÿ)^๐Ÿ/|๐’™ + ๐Ÿ| ใ€— +๐‘ช

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