Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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  1. Chapter 7 Class 12 Integrals
  2. 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 + (โˆ’4๐‘ก โˆ’10)/(๐‘ก^2 + 7๐‘ก + 12) = 1 + (โˆ’(4๐‘ก + 10))/(๐‘ก + 3)(๐‘ก + 4) Rough = 1 โˆ’ ( (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‘กโˆ’10 = ๐ด(๐‘ก+4)+๐ต(๐‘ก+3) Putting t = โˆ’ 4 in (1) 4(โˆ’4)+10 = ๐ด(โˆ’4+4)+๐ต(โˆ’4+3) โˆ’16+10 = ๐ดร—0+๐ต(โˆ’1) โ€ฆ(1) โˆ’6 = ๐ดร—0+๐ต(โˆ’1) โˆ’6 = โˆ’๐ต ๐ต = 6 Putting t = โˆ’3 in (1) 4๐‘กโˆ’10 = ๐ด(๐‘ก+4)+๐ต(๐‘ก+3) 4(โˆ’3)+10 = ๐ด(โˆ’3+4)+๐ต(โˆ’3+3) โˆ’12+10 = ๐ดร—1+๐ตร—0 โˆ’2 = ๐ด ๐ด = โˆ’2 Hence we can write (4๐‘ก + 10)/((๐‘ก + 3) (๐‘ก + 4) ) = (โˆ’2)/((๐‘ก + 3) ) + 6/((๐‘ก + 4) ) Putting back t = ๐‘ฅ^2 (4๐‘ฅ^2 โˆ’ 10)/((๐‘ฅ^2 + 3) (๐‘ฅ^2 + 4) ) = (โˆ’2)/((๐‘ฅ^2 + 3) ) + 6/((๐‘ฅ^2 + 4) ) Therefore โˆซ1โ–’(๐‘ฅ2+ 1)(๐‘ฅ2+ 2)/(๐‘ฅ2+ 3)(๐‘ฅ2+ 4) = โˆซ1โ–’ใ€–1โˆ’[(โˆ’2)/((๐‘ฅ^2 + 3) ) + 6/((๐‘ฅ^2 + 4) )] ใ€— ๐‘‘๐‘ฅ = โˆซ1โ–’1. ๐‘‘๐‘ฅ + โˆซ1โ–’2/((๐‘ฅ^2 + 3) ) ๐‘‘๐‘ฅ โˆ’ โˆซ1โ–’6/((๐‘ฅ^2 + 4) ) ๐‘‘๐‘ฅ = โˆซ1โ–’1. ๐‘‘๐‘ฅ + 2โˆซ1โ–’1/(๐‘ฅ^2 + (โˆš3)^2 ) ๐‘‘๐‘ฅ โˆ’ 6โˆซ1โ–’1/((๐‘ฅ^2 +2^2 ) ) ๐‘‘๐‘ฅ = ๐‘ฅ + 2 ร— 1/โˆš3 tan^(โˆ’1)โกใ€– ๐‘ฅ/โˆš3ใ€— โˆ’ 6 ร— 1/2 tan^(โˆ’1)โกใ€– ๐‘ฅ/2ใ€—+๐ถ = ๐’™ + ๐Ÿ/โˆš๐Ÿ‘ ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โก(๐’™/โˆš๐Ÿ‘)โˆ’๐Ÿ‘ ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โก(๐’™/๐Ÿ)+๐‘ช

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.