Integration by partial fraction - Type 1

Chapter 7 Class 12 Integrals
Concept wise

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Ex 7.5, 2 1/(๐ฅ2โ 9) Solving integrand 1/(๐ฅ2โ 9)=1/((๐ฅ โ 3) (๐ฅ + 3) ) We can write it as 1/((๐ฅ โ 3) (๐ฅ + 3) )=๐ด/((๐ฅ โ 3) ) + ๐ต/((๐ฅ + 3) ) 1/((๐ฅ โ 3) (๐ฅ + 3) )=(๐ด(๐ฅ + 3) + ๐ต(๐ฅ โ 3))/((๐ฅ โ 3) (๐ฅ + 3) ) " " Cancelling denominator 1 = ๐ด(๐ฅ + 3) + ๐ต(๐ฅ โ 3) Putting ๐ฅ = 3 in (1) 1 = ๐ด(๐ฅ+3)+๐ต(๐ฅโ3) 1 = ๐ด(3+3) + ๐ต(3โ3) 1 = ๐ดร6+ ๐ตร0 1 = 6๐ด ๐ด = 1/6 Similarly Putting y=โ3 in (1) 1 = ๐ด(โ3+3) + ๐ต(โ3โ3) 1 = ๐ดร0+ ๐ตร(โ6) 1 = โ6๐ต ๐ต = (โ1)/6 Hence we can write it as 1/((๐ฅ โ 3) (๐ฅ + 3) ) = 1/6(๐ฅ โ 3) โ 1/6(๐ฅ + 3) Therefore โซ1โ1/((๐ฅ โ 3) (๐ฅ + 3) ) ๐๐ฅ = 1/6 โซ1โ1/((๐ฅ โ 3) ) ๐๐ฅ โ 1/6 โซ1โ1/((๐ฅ + 3) ) ๐๐ฅ = 1/6 ใlog ใโก|๐ฅโ3|โ 1/6 ใlog ใโก|๐ฅ+3|+๐ถ = 1/6 (ใlog ใโก|๐ฅโ3|โใlog ใโก|๐ฅ+3| )+๐ถ = ๐/๐ ใ๐๐๐ ใโก|(๐ โ ๐)/(๐ + ๐)|+๐ช