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Find : ∫(2x^2+3)/(x^2 (x^2+9) ) dx;x≠0

This question is similar to Ex 7.5, 18 Chapter 7 Class 12

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Transcript

(2๐‘ฅ^2 + 3)/(๐‘ฅ^2 (๐‘ฅ^2 + 9) ) Let t = ๐’™^๐Ÿ = (2๐‘ก + 3)/๐‘ก(๐‘ก + 9) We can write (2๐‘ก + 3)/๐‘ก(๐‘ก + 9) = ๐‘จ/๐’• + ๐‘ฉ/((๐’• + ๐Ÿ—) ) (2๐‘ก + 3)/๐‘ก(๐‘ก + 9) = (๐ด(๐‘ก + 9) +๐ต๐‘ก)/(๐‘ก(๐‘ก + 9) ) Cancelling denominator ๐Ÿ๐’•+๐Ÿ‘ = ๐‘จ(๐’•+๐Ÿ—)+๐‘ฉ๐’• Putting t = โˆ’๐Ÿ— in (1) 2(โˆ’9)+3 = ๐ด(โˆ’9+9)+๐ต(โˆ’9) โˆ’18+3 = ๐ดร—0+๐ต(โˆ’9) โˆ’15 = 0+๐ต(โˆ’9) โˆ’15 = โˆ’9๐ต B = 15/9 ๐‘ฉ = ๐Ÿ“/๐Ÿ‘ Putting t = ๐ŸŽ in (1) 2(0)+3 = ๐ด(0+9)+๐ต(0) 3 = 9๐ด+0 A = 9/3 A = ๐Ÿ/๐Ÿ‘ Hence we can write (2๐‘ก + 3)/๐‘ก(๐‘ก + 9) = (1/3)/๐‘ก + (5/3)/((๐‘ก + 9 ) ) (2๐‘ก + 3)/๐‘ก(๐‘ก + 9) = 1/3๐‘ก + 5/(3(๐‘ก + 9)) Putting back t = ๐’™^๐Ÿ (๐Ÿ๐’™^๐Ÿ+ ๐Ÿ‘)/(๐’™^๐Ÿ (๐’™^๐Ÿ + ๐Ÿ—) ) = ๐Ÿ/(๐Ÿ‘๐’™^๐Ÿ ) + ๐Ÿ“/(๐Ÿ‘(๐’™^(๐Ÿ )+ ๐Ÿ—)) Therefore, โˆซ1โ–’(2๐‘ฅ^2 + 3)/(๐‘ฅ^2 (๐‘ฅ^2 +9)) ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–1/(3๐‘ฅ^2 ) ๐‘‘๐‘ฅ"+ " โˆซ1โ–’5/(3(๐‘ฅ^2+ 9))ใ€— ๐‘‘๐‘ฅ = 1/3 โˆซ1โ–’ใ€–1/๐‘ฅ^2 ๐‘‘๐‘ฅ" + " 5/3 โˆซ1โ–’1/((๐‘ฅ^2+ 9))ใ€— ๐‘‘๐‘ฅ = ๐Ÿ/๐Ÿ‘ โˆซ1โ–’ใ€–๐Ÿ/๐’™^๐Ÿ ๐’…๐’™" + " ๐Ÿ“/๐Ÿ‘ โˆซ1โ–’๐Ÿ/((๐’™^๐Ÿ+ ใ€–(๐Ÿ‘)ใ€—^๐Ÿ))ใ€— ๐’…๐’™ = 1/3 ร— (โˆ’๐Ÿ)/๐’™ + 5/3 ร— ๐Ÿ/๐Ÿ‘ ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ)โกใ€– ๐’™/๐Ÿ‘ใ€— + C = (โˆ’๐Ÿ)/๐Ÿ‘๐’™ + ๐Ÿ“/๐Ÿ— ใ€–๐’•๐’‚๐’ใ€—^(โˆ’๐Ÿ) (๐’™/๐Ÿ‘) + C

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.