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Now learn Economics at Teachoo for Class 12


Transcript

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’™/((๐’™ + ๐Ÿ) (๐’™ + ๐Ÿ) ) " " = ๐‘จ/((๐’™ + ๐Ÿ) ) + ๐‘ฉ/((๐’™ + ๐Ÿ) ) 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’™ = ๐‘จ(๐’™+๐Ÿ) + ๐‘ฉ(๐’™+๐Ÿ) Putting ๐’™=โˆ’๐Ÿ in (1) โˆ’1=๐ด(โˆ’1+2) + ๐ต(โˆ’1+1) โˆ’1=๐ด ร— 1+ ๐ต ร— 0 โˆ’1=๐ด ๐‘จ=โˆ’๐Ÿ Putting ๐’™=โˆ’๐Ÿ in (1) โˆ’2 = ๐ด(โˆ’2+2) + ๐ต(โˆ’2+1) โˆ’2 = ๐ด ร— 0+ ๐ต ร— (โˆ’1) โˆ’2 = โˆ’ ๐ต ๐‘ฉ = ๐Ÿ Hence we can write it as ๐’™/((๐’™ + ๐Ÿ) (๐’™ + ๐Ÿ) ) = (โˆ’๐Ÿ)/((๐’™ + ๐Ÿ) ) + ๐Ÿ/((๐’™ + ๐Ÿ) ) Therefore โˆซ1โ–’๐’™/((๐’™ + ๐Ÿ) (๐’™ + ๐Ÿ) ) ๐’…๐’™ = 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€–๐ฅ๐จ๐  ใ€—โก|๐’™+๐Ÿ|+๐Ÿ ใ€–๐ฅ๐จ๐  ใ€—โก|๐’™+๐Ÿ|+๐‘ช = 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 As ๐’๐’๐’ˆ ๐‘จโˆ’๐’๐’๐’ˆ ๐‘ฉ=๐‘™๐‘œ๐‘” ๐ด/๐ต As (๐‘ฅ+2)^2is always positive

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.