Ex 7.5

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.5, 20 Integrate the function 1/(๐ฅ(๐ฅ4โ1) ) 1/(๐ฅ(๐ฅ4 โ 1) ) Multiplying integrand by ๐ฅ^3/๐ฅ^3 = 1/(๐ฅ(๐ฅ^4 โ 1) ) ร ๐ฅ^3/๐ฅ^3 = ๐ฅ^3/(๐ฅ^4 (๐ฅ^4 โ 1) ) Let t = ๐ฅ^4 Differentiating both sides ๐ค.๐.๐ก.๐ฅ ๐๐ก/๐๐ฅ = 4๐ฅ^3 ๐๐ก/(4๐ฅ^3 ) = ๐๐ฅ Substituting value of ๐ก = ๐ฅ^4 & ๐๐ฅ = ๐๐ก/(4๐ฅ^3 ) " " โซ1โ๐ฅ^3/(๐ฅ^4 (๐ฅ^4โ 1) ) ๐๐ฅ = โซ1โ๐ฅ^3/(๐ก(๐ก โ 1) ) ๐๐ก/(4๐ฅ^3 ) " " = 1/4 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ต ๐ก)/๐ก(๐ก โ 1) Cancelling denominator 1 = ๐ด(๐กโ1)+๐ต๐ก โฆ(1) 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 in (1) 1 = A(tโ1)+Bt 1 = ๐ด(1โ1)+๐ตร1 1 = ๐ดร0+๐ต 1 = ๐ต ๐ต = 1 Therefore 1/4 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 =ใ ๐ฅใ^4 = ๐/๐ ใ๐ฅ๐จ๐  ใโก|(๐^๐ โ ๐)/๐^๐ |+๐ช ("As " ๐๐๐ ๐ดโ๐๐๐ ๐ต" = " ๐๐๐ ๐ด/๐ต)