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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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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’™/((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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.