Integration by partial fraction - Type 1

Chapter 7 Class 12 Integrals
Concept 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/((๐ + ๐) (๐ + ๐) ) " " = ๐จ/((๐ + ๐) ) + ๐ฉ/((๐ + ๐) ) 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) + ๐ฉ(๐+๐) Putting ๐=โ๐ in (1) โ1=๐ด(โ1+2) + ๐ต(โ1+1) โ1=๐ด ร 1+ ๐ต ร 0 โ1=๐ด ๐จ=โ๐ Putting ๐=โ๐ in (1) โ2 = ๐ด(โ2+2) + ๐ต(โ2+1) โ2 = ๐ด ร 0+ ๐ต ร (โ1) โ2 = โ ๐ต ๐ฉ = ๐ Hence we can write it as ๐/((๐ + ๐) (๐ + ๐) ) = (โ๐)/((๐ + ๐) ) + ๐/((๐ + ๐) ) Therefore โซ1โ๐/((๐ + ๐) (๐ + ๐) ) ๐๐ = โซ1โ(โ1)/((๐ฅ + 1) ) ๐๐ฅ + โซ1โ2/((๐ฅ + 2) ) ๐๐ฅ = โ1โซ1โ1/((๐ฅ + 1) ) ๐๐ฅ + 2โซ1โ1/((๐ฅ + 2) ) ๐๐ฅ = โใ๐ฅ๐จ๐  ใโก|๐+๐|+๐ ใ๐ฅ๐จ๐  ใโก|๐+๐|+๐ช = โ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)^๐/|๐ + ๐| ใ +๐ช As ๐๐๐ ๐จโ๐๐๐ ๐ฉ=๐๐๐ ๐ด/๐ต As (๐ฅ+2)^2is always positive