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 7.9, 16 โˆซ_1^2โ–’(5๐‘ฅ^2)/(๐‘ฅ^2 + 4๐‘ฅ + 3) ๐‘‘๐‘ฅ Let F(๐‘ฅ)=โˆซ1โ–’ใ€–(5๐‘ฅ^2)/(๐‘ฅ^2 + 4๐‘ฅ + 3) ๐‘‘๐‘ฅใ€— =5โˆซ1โ–’ใ€–๐‘ฅ^2/(๐‘ฅ^2 + 4๐‘ฅ + 3) ๐‘‘๐‘ฅใ€— =5โˆซ1โ–’ใ€–(๐‘ฅ^2 + 4๐‘ฅ + 3 โˆ’ 4๐‘ฅ โˆ’ 3)/(๐‘ฅ^2 + 4๐‘ฅ + 3) ๐‘‘๐‘ฅใ€— =5โˆซ1โ–’ใ€–(๐‘ฅ^2 + 4๐‘ฅ + 3)/(๐‘ฅ^2 + 4๐‘ฅ + 3) ๐‘‘๐‘ฅใ€—โˆ’5โˆซ1โ–’ใ€–( (4๐‘ฅ + 3))/(๐‘ฅ^2 + 4๐‘ฅ + 3) ๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–(5โˆ’(20๐‘ฅ + 15 )/(๐‘ฅ^2 + 4๐‘ฅ + 3)) ๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–(5โˆ’(20๐‘ฅ + 15 )/(๐‘ฅ^2 + 3๐‘ฅ + ๐‘ฅ + 3)) ๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–(5โˆ’(20๐‘ฅ + 15 )/(๐‘ฅ(๐‘ฅ + 3) + 1(๐‘ฅ + 3))) ๐‘‘๐‘ฅใ€— =โˆซ1โ–’ใ€–(5โˆ’(20๐‘ฅ + 15 )/((๐‘ฅ + 3) (๐‘ฅ + 1))) ๐‘‘๐‘ฅใ€— Now, Let (20๐‘ฅ + 15)/(๐‘ฅ + 3)(๐‘ฅ + 1) =A/(๐‘ฅ + 3)+B/(๐‘ฅ + 1) (20๐‘ฅ + 15)/(๐‘ฅ + 3)(๐‘ฅ + 1) =(A(๐‘ฅ + 1) + B(๐‘ฅ + 3))/(๐‘ฅ + 3)(๐‘ฅ + 1) Canceling denominator 20๐‘ฅ+15=A(๐‘ฅ + 1) + B(๐‘ฅ + 3) Putting ๐‘ฅ=โˆ’1 20(โˆ’1)+15=A(โˆ’1 + 1) + B(โˆ’1 + 3) โˆ’20+15=Aร—0+B (2) Putting ๐‘ฅ=โˆ’1 20(โˆ’1)+15=A(โˆ’1 + 1) + B(โˆ’1 + 3) โˆ’20+15=Aร—0+B (2) โˆ’5=2B B=(โˆ’5)/( 2) Similarly Putting ๐‘ฅ=โˆ’3 20(โˆ’3)+15=A(โˆ’3+1)+B(โˆ’3+3) โˆ’60+15=A(โˆ’2) Bร—0 โˆ’45=โˆ’2A A=45/2 Hence โˆซ1โ–’โ–ˆ((5๐‘ฅ^2)/(๐‘ฅ^2 + 4๐‘ฅ + 3) " " =โˆซ1โ–’ใ€–(5โˆ’(20๐‘ฅ + 15 )/(๐‘ฅ^2 + 4๐‘ฅ + 3)) ๐‘‘๐‘ฅใ€—) =โˆซ1โ–’ใ€–5โˆ’A/(๐‘ฅ+3)โˆ’ใ€— B/(๐‘ฅ+1) ๐‘‘๐‘ฅ =โˆซ1โ–’ใ€–5 ๐‘‘๐‘ฅใ€—โˆ’โˆซ1โ–’ใ€–(45/2)/(๐‘ฅ +_3) ๐‘‘๐‘ฅโˆ’โˆซ1โ–’ใ€–(((โˆ’5)/( 2)))/(๐‘ฅ+1) ๐‘‘๐‘ฅใ€—ใ€— =5๐‘ฅโˆ’45/2 ๐‘™๐‘œ๐‘”|๐‘ฅ+3|+5/2 ๐‘™๐‘œ๐‘”|๐‘ฅ+1| Hence F(๐‘ฅ)=5๐‘ฅโˆ’5/2 [9 ๐‘™๐‘œ๐‘”|๐‘ฅ+3|โˆ’๐‘™๐‘œ๐‘”|๐‘ฅ+1|] Now, โˆซ_1^2โ–’ใ€–(5๐‘ฅ^2)/(๐‘ฅ^2 + 4๐‘ฅ + 3) ๐‘‘๐‘ฅ=๐น(2)โˆ’๐น(1) ใ€— =[5ร—2โˆ’5/2 (9 ๐‘™๐‘œ๐‘”|2+3|โˆ’๐‘™๐‘œ๐‘”|2+1|)] โˆ’ [5 ร—1โˆ’5/2 (9 ๐‘™๐‘œ๐‘”|1+3|โˆ’๐‘™๐‘œ๐‘”|1+1|)] =10โˆ’5/2 [9 ๐‘™๐‘œ๐‘” 5โˆ’๐‘™๐‘œ๐‘” 3]โˆ’5+5/2 [9 ๐‘™๐‘œ๐‘” 4โˆ’๐‘™๐‘œ๐‘” 2] =10โˆ’5โˆ’5/2 [9๐‘™๐‘œ๐‘” 5โˆ’๐‘™๐‘œ๐‘” 3โˆ’9๐‘™๐‘œ๐‘” 4+๐‘™๐‘œ๐‘” 2)] =10โˆ’5โˆ’5/2 [9๐‘™๐‘œ๐‘” 5โˆ’9๐‘™๐‘œ๐‘” 4โˆ’(๐‘™๐‘œ๐‘” 3โˆ’๐‘™๐‘œ๐‘” 2)] =๐Ÿ“โˆ’๐Ÿ“/๐Ÿ (๐Ÿ— ๐ฅ๐จ๐  ๐Ÿ“/๐Ÿ’โˆ’๐ฅ๐จ๐  ๐Ÿ‘/๐Ÿ) (log a โˆ’ log b = log ๐‘Ž/๐‘)

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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.