Integration by partial fraction - Type 1

Chapter 7 Class 12 Integrals
Concept wise

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Transcript

Ex 7.5, 21 Integrate the function 1/((๐^๐ฅ โ 1) ) [Hint : Put ex = t] Let ๐^๐ฅ = ๐ก Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐^๐ฅ = ๐๐ก/๐๐ฅ ๐๐ฅ = ๐๐ก/๐^๐ฅ Therefore โซ1โ1/((๐^๐ฅ โ 1) ) ๐๐ฅ = โซ1โ1/((๐ก โ 1) ) ๐๐ก/๐^๐ฅ = โซ1โ๐๐ก/(๐ก(๐ก โ 1) ) We can write integrand as 1/(๐ก(๐ก โ 1) ) = ๐ด/๐ก + ๐ต/(๐ก โ 1) 1/(๐ก(๐ก โ 1) ) = (๐ด(๐ก โ 1) + ๐ต๐ก)/๐ก(๐ก โ 1) Cancelling denominator 1 = ๐ด(๐กโ1)+๐ต๐ก Putting t = 0 in (1) 1 = ๐ด(0โ1)+๐ตร0 1 = ๐ดร(โ1) 1 = โ๐ด ๐ด = โ1 Putting t = 1 1 = ๐ด(1โ1)+๐ตร1 1 = ๐ดร0+๐ต 1 = ๐ต ๐ต = 1 Therefore โซ1โ1/(๐ก(๐ก โ 1) ) ๐๐ก = โซ1โ(โ1)/(๐ก ) ๐๐ก + โซ1โ1/(๐ก โ 1 ) = โใlog ใโก|๐ก|+ใlog ใโก|๐กโ1|+๐ถ = ใlog ใโก|(๐ก โ 1)/๐ก|+๐ถ Putting back t = ๐^๐ฅ = ใ๐๐๐ ใโก|(๐^๐ฅ โ 1)/๐^๐ฅ |+๐ถ = ใ๐ฅ๐จ๐  ใโก((๐^๐ โ ๐)/๐^๐ )+๐ช Since ex > 1 for x > 0 โด ex โ 1 > 0 โ |(๐^๐ โ ๐)/๐^๐ |=((๐^๐ โ ๐)/๐^๐ ) ("As " ๐๐๐ ๐ดโ๐๐๐ ๐ต" = " ๐๐๐ ๐ด/๐ต)