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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Example 16 Find โˆซ1โ–’(๐‘ฅ^2+ ๐‘ฅ +1 ๐‘‘๐‘ฅ )/((๐‘ฅ + 2) (๐‘ฅ^2+1) ) We can write equation as (๐‘ฅ^2+ ๐‘ฅ + 1)/((๐‘ฅ + 1) (๐‘ฅ + 2) )=๐ด/(๐‘ฅ + 2) + (๐ต๐‘ฅ + ๐ถ)/(๐‘ฅ^2+ 1) Cancelling denominator ใ€– ๐‘ฅใ€—^2+ ๐‘ฅ+1=๐ด(๐‘ฅ^2+1)+(๐ต๐‘ฅ+๐ถ) (๐‘ฅ+2) Putting x = โˆ’๐Ÿ (โˆ’2)^2+(โˆ’2)+1=๐ด((โˆ’2)^2+1)+0 4โˆ’2+1= 5A 3/5 = A Putting x = ๐ŸŽ ๐‘ฅ^2+ ๐‘ฅ+1=๐ด(๐‘ฅ^2+1)+(๐ต๐‘ฅ+๐ถ) (๐‘ฅ+2) 0+0+0= A(0 + 1) + (0 + C) (0 + 2) 1 = A + 2C 1 = 3/5 + 2C 1 โ€“ 3/5 = 2C 2/5 = 2C C = 1/5 Putting x = 1 ๐‘ฅ^2+ ๐‘ฅ+1=๐ด(๐‘ฅ^2+1)+(๐ต๐‘ฅ+๐ถ) (๐‘ฅ+2) 1+1+1= 2A + (B + C)(3) 3 = 2A + 3 (B + C) 3 = 2(3/5) + 3 (B+1/5) 3 โ€“ 6/5 = 3 (B+1/5) 9/5 = 3 (B+1/5) 3/5 โ€“ 1/5 = B B = 2/5 Thus, (๐‘ฅ^2+ ๐‘ฅ + 1)/((๐‘ฅ + 1) (๐‘ฅ + 2) )=๐ด/(๐‘ฅ + 2) + (๐ต๐‘ฅ + ๐ถ)/(๐‘ฅ^2+ 1) (๐‘ฅ^2+ ๐‘ฅ + 1)/((๐‘ฅ + 1)(๐‘ฅ^2+ 1)) = 3/(5 (๐‘ฅ + 2)) + (1 (2๐‘ฅ + 1))/(5 (๐‘ฅ^2 + 1)) Hence, our equation becomes โˆซ1โ–’(๐‘ฅ^2+ ๐‘ฅ + 1)/((๐‘ฅ + 2) (๐‘ฅ^2 + 1)) ๐‘‘๐‘ฅ= โˆซ1โ–’3/(5(๐‘ฅ^2 + 1)) ๐‘‘๐‘ฅ+โˆซ1โ–’1/5 ((2๐‘ฅ + 1))/(๐‘ฅ^2 + 1) ๐‘‘๐‘ฅ = โˆซ1โ–’3/(5(๐‘ฅ^2 + 1)) ๐‘‘๐‘ฅ+ 1/5 โˆซ1โ–’ใ€–2๐‘ฅ/(๐‘ฅ^2 + 1) ๐‘‘๐‘ฅ+ใ€— 1/5 โˆซ1โ–’1/(๐‘ฅ^2 + 1) ๐‘‘๐‘ฅ ๐ˆ๐Ÿ 1/5 โˆซ1โ–’2๐‘ฅ/(๐‘ฅ^2+ 1) ๐‘‘๐‘ฅ Let ๐‘ก=๐‘ฅ^2+ 1 ๐‘‘๐‘ก/๐‘‘๐‘ฅ=2๐‘ฅ ๐‘‘๐‘ก=2๐‘ฅ ๐‘‘๐‘ฅ Substituting, =1/5 โˆซ1โ–’๐‘‘๐‘ก/๐‘ก = 1/5 log |๐‘ก| + C_2 = 1/5 log |๐‘ฅ^2+1| + C_2 ๐ˆ๐Ÿ‘ 1/5 โˆซ1โ–’1/(๐‘ฅ^2+ 1) ๐‘‘๐‘ฅ = 1/5 ใ€–๐‘ก๐‘Ž๐‘›ใ€—^(โˆ’1) (๐‘ฅ)+C_3 Hence โˆซ1โ–’(๐‘ฅ^2+ ๐‘ฅ + 1)/((๐‘ฅ + 2) (๐‘ฅ^2+ 1)) ๐‘‘๐‘ฅ =๐Ÿ‘/๐Ÿ“ ๐’๐’๐’ˆ|๐’™+๐Ÿ|+๐Ÿ/๐Ÿ“ ๐’๐’๐’ˆ|๐’™^๐Ÿ+๐Ÿ|+๐Ÿ/๐Ÿ“ ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ) (๐’™)+ C where C = C_1+ C_2+C_3

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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.