Ex 7.5

Chapter 7 Class 12 Integrals
Serial order wise

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Ex 7.5, 14 (3๐ฅ โ 1)/(๐ฅ + 2)^2 We can write it as (3๐ฅ โ 1)/(๐ฅ + 2)^2 = ๐ด/((๐ฅ + 2) )+๐ต/(๐ฅ + 2)^2 (3๐ฅ โ 1)/(๐ฅ + 2)^2 = (๐ด(๐ฅ + 2) + ๐ต)/(๐ฅ + 2)^2 By cancelling denominator 3๐ฅโ1 = ๐ด(๐ฅ + 2)+๐ต Putting x = โ2 in (1) 3๐ฅโ1=๐ด (๐ฅ+2)+๐ต 3(โ2)โ1=๐ด (โ2+2)+๐ต โ6โ1=๐ดร0+๐ต โ7=๐ต ๐ต=โ7 Putting x = 0 in (1) 3๐ฅโ1=๐ด (๐ฅ+2)+๐ต 3ร0โ1=๐ด (0+2)+๐ต โ1=2๐ด+๐ต โ1=2๐ด+(โ7) โ1+7=2๐ด 2๐ด=6 ๐ด=3 Hence we can write it as (3๐ฅ โ 1)/(๐ฅ + 2)^2 = 3/( ๐ฅ + 2)+(โ7)/(๐ฅ + 2)^2 Integrating ๐ค.๐.๐ก.๐ฅ. โซ1โ(3๐ฅ โ 1)/(๐ฅ + 2)^2 ๐๐ฅ=โซ1โ(3/(๐ฅ + 2) โ 5/(๐ฅ + 2)^2 ) ๐๐ฅ =3โซ1โใ1/(๐ฅ + 2) ใ ๐๐ฅโ7โซ1โใ1/(๐ฅ + 2)^2 ใ ๐๐ฅ =3 ใlog ใโก|๐ฅ+2|โ7โซ1โใ(๐ฅ+2)^(โ2) ใ ๐๐ฅ =3 ใlog ใโก|๐ฅ+2|โ7 (๐ฅ + 2)^(โ2 + 1)/(โ2 + 1) =3 ใlog ใโก|๐ฅ+2|โ7 (๐ฅ + 2)^(โ1)/(โ1) +๐ถ =๐ ใ๐ฅ๐จ๐  ใโก|๐+๐|+๐/((๐ + ๐) ) +๐ช