Ex 7.5

Chapter 7 Class 12 Integrals
Serial order wise

Transcript

Ex 7.5, 18 Integrate the function (๐ฅ2+ 1)(๐ฅ2+ 2)/(๐ฅ2+ 3)(๐ฅ2+ 4) (๐ฅ^2 + 1)(๐ฅ^2 + 2)/(๐ฅ^2 + 3)(๐ฅ^2 + 4) " " Let t = ๐ฅ^2 = (๐ก + 1)(๐ก + 2)/(๐ก + 3)(๐ก + 4) = (๐ก^2 + 3๐ก + 2)/(๐ก^2 + 7๐ก + 12) = 1 + (โ4๐ก โ10)/(๐ก^2 + 7๐ก + 12) = 1 + (โ(4๐ก + 10))/(๐ก + 3)(๐ก + 4) Rough = 1 โ ( (4๐ก + 10))/(๐ก + 3)(๐ก + 4) We can write (4๐ก + 10)/((๐ก + 3) (๐ก + 4) ) = ๐ด/((๐ก + 3) ) + ๐ต/((๐ก + 4) ) (4๐ก + 10)/((๐ก + 3) (๐ก + 4) ) = (๐ด(๐ก + 4) + ๐ต(๐ก + 3))/((๐ก + 3) (๐ก + 4) ) Cancelling denominator 4๐กโ10 = ๐ด(๐ก+4)+๐ต(๐ก+3) Putting t = โ 4 in (1) 4(โ4)+10 = ๐ด(โ4+4)+๐ต(โ4+3) โ16+10 = ๐ดร0+๐ต(โ1) โฆ(1) โ6 = ๐ดร0+๐ต(โ1) โ6 = โ๐ต ๐ต = 6 Putting t = โ3 in (1) 4๐กโ10 = ๐ด(๐ก+4)+๐ต(๐ก+3) 4(โ3)+10 = ๐ด(โ3+4)+๐ต(โ3+3) โ12+10 = ๐ดร1+๐ตร0 โ2 = ๐ด ๐ด = โ2 Hence we can write (4๐ก + 10)/((๐ก + 3) (๐ก + 4) ) = (โ2)/((๐ก + 3) ) + 6/((๐ก + 4) ) Putting back t = ๐ฅ^2 (4๐ฅ^2 โ 10)/((๐ฅ^2 + 3) (๐ฅ^2 + 4) ) = (โ2)/((๐ฅ^2 + 3) ) + 6/((๐ฅ^2 + 4) ) Therefore โซ1โ(๐ฅ2+ 1)(๐ฅ2+ 2)/(๐ฅ2+ 3)(๐ฅ2+ 4) = โซ1โใ1โ[(โ2)/((๐ฅ^2 + 3) ) + 6/((๐ฅ^2 + 4) )] ใ ๐๐ฅ = โซ1โ1. ๐๐ฅ + โซ1โ2/((๐ฅ^2 + 3) ) ๐๐ฅ โ โซ1โ6/((๐ฅ^2 + 4) ) ๐๐ฅ = โซ1โ1. ๐๐ฅ + 2โซ1โ1/(๐ฅ^2 + (โ3)^2 ) ๐๐ฅ โ 6โซ1โ1/((๐ฅ^2 +2^2 ) ) ๐๐ฅ = ๐ฅ + 2 ร 1/โ3 tan^(โ1)โกใ ๐ฅ/โ3ใ โ 6 ร 1/2 tan^(โ1)โกใ ๐ฅ/2ใ+๐ถ = ๐ + ๐/โ๐ ใ๐๐๐ใ^(โ๐)โก(๐/โ๐)โ๐ ใ๐๐๐ใ^(โ๐)โก(๐/๐)+๐ช