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

Misc 24 Integrate the function (โˆš(๐‘ฅ^2 + 1) [logโกใ€–(๐‘ฅ^2+ 1) โˆ’ 2 logโก๐‘ฅ ใ€— ] )/๐‘ฅ^4 โˆซ1โ–’(โˆš(๐‘ฅ^2 + 1) [logโกใ€–(๐‘ฅ^2+ 1) โˆ’2 logโก๐‘ฅ ใ€— ] )/๐‘ฅ^4 ๐‘‘๐‘ฅ Taking ๐‘ฅ^2common from โˆš(๐‘ฅ^2+1) = โˆซ1โ–’(ใ€–ใ€–(๐‘ฅใ€—^2) ใ€—^(1/2) (1 + 1/๐‘ฅ^2 )^(1/2) (logโกใ€–(๐‘ฅ^2+1)ใ€— โˆ’ logโกใ€–๐‘ฅ^2 ใ€— ))/๐‘ฅ^4 ๐‘‘๐‘ฅ = โˆซ1โ–’(๐‘ฅ (1+ 1/๐‘ฅ^2 )^(1/2) (logโกใ€– ((๐‘ฅ^(2 )+ 1))/๐‘ฅ^2 ใ€— ))/๐‘ฅ^4 ๐‘‘๐‘ฅ (๐‘› logโก๐‘š ใ€–=logใ€—โกใ€–๐‘š^๐‘› ใ€— " " ) (logโก๐‘šโˆ’logโก๐‘› ใ€–=logใ€—โกใ€–๐‘š/๐‘›ใ€— " " ) = โˆซ1โ–’( (1+ 1/๐‘ฅ^2 )^(1/2) (logโก(1 + 1/๐‘ฅ^2 ) ))/๐‘ฅ^3 Let t = 1 + 1/๐‘ฅ^2 ๐‘‘๐‘ก/๐‘‘๐‘ฅ=(โˆ’2)/๐‘ฅ^3 (โˆ’1)/2 ๐‘‘๐‘ก=๐‘‘๐‘ฅ/๐‘ฅ^3 Substituting, = โˆ’1/2 โˆซ1โ–’๐‘ก^(1/2) ใ€– log ๐‘กใ€—โกใ€– ๐‘‘๐‘กใ€— Now we know, โˆซ1โ–’ใ€–๐‘“(๐‘ฅ)๐‘”(๐‘ฅ)๐‘‘๐‘ฅ=๐‘“(๐‘ฅ)โˆซ1โ–’ใ€–๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–(๐‘“^โ€ฒ (๐‘ฅ) โˆซ1โ–’๐‘”(๐‘ฅ)๐‘‘๐‘ฅ) ๐‘‘๐‘ฅใ€—ใ€—ใ€— Put f(t) = log t & g(t) = t 1/2 = โˆซ1โ–’( (1+ 1/๐‘ฅ^2 )^(1/2) (logโก(1 + 1/๐‘ฅ^2 ) ))/๐‘ฅ^3 Let t = 1 + 1/๐‘ฅ^2 ๐‘‘๐‘ก/๐‘‘๐‘ฅ=(โˆ’2)/๐‘ฅ^3 (โˆ’1)/2 ๐‘‘๐‘ก=๐‘‘๐‘ฅ/๐‘ฅ^3 Substituting value of t and dt = (โˆ’1)/2 โˆซ1โ–’๐‘ก^(1/2) ใ€– log ๐‘กใ€—โกใ€– ๐‘‘๐‘กใ€— Now we know, โˆซ1โ–’ใ€–๐‘“(๐‘ฅ)๐‘”(๐‘ฅ)๐‘‘๐‘ฅ=๐‘“(๐‘ฅ)โˆซ1โ–’ใ€–๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–(๐‘“^โ€ฒ (๐‘ฅ) โˆซ1โ–’๐‘”(๐‘ฅ)๐‘‘๐‘ฅ) ๐‘‘๐‘ฅใ€—ใ€—ใ€— Put f(t) = log t & g(t) = ๐‘ก^(1/2) Hence, (โˆ’1)/2 โˆซ1โ–’ใ€–๐‘ก^(1/2) logโกใ€–๐‘ก ๐‘‘๐‘ก=(โˆ’1)/2 (logโกใ€–๐‘ก โˆซ1โ–’ใ€–๐‘ก^(1/2) ๐‘‘๐‘กใ€—โˆ’โˆซ1โ–’((๐‘‘(logโกใ€–๐‘ก)ใ€—)/๐‘‘๐‘ก โˆซ1โ–’๐‘ก^(1/2) ๐‘‘๐‘ก) ๐‘‘๐‘กใ€— )ใ€— ใ€— = (โˆ’1)/2 (logโกใ€–๐‘ก (๐‘ก^(3/2)/(3/2))โˆ’โˆซ1โ–’ใ€–1/๐‘กร—(๐‘ก^(3/2)/(3/2)) ใ€—ใ€— ๐‘‘๐‘ก) = (โˆ’1)/2 (2/3 ๐‘ก^(3/2) logโกใ€–๐‘กโˆ’2/3ใ€— โˆซ1โ–’ใ€–๐‘ก^(1/2) ๐‘‘๐‘กใ€—) = (โˆ’1)/2 (2/3 ๐‘ก^(3/2) logโกใ€–๐‘กโˆ’2/3ใ€— ( ใ€–2๐‘กใ€—^(3/2))/3) = (โˆ’1)/3 ๐‘ก^(3/2) logโก๐‘ก + 2/9 ๐‘ก^(3/2) Putting value of t = 1 + 1/๐‘ฅ^2 = (โˆ’1)/3 (1+1/๐‘ฅ^2 )^(3/2) logโกใ€–(1+1/๐‘ฅ^2 )+2/9 " " (1+1/๐‘ฅ^2 )^(3/2)+ใ€— C = (โˆ’๐Ÿ)/๐Ÿ‘ (๐Ÿ+๐Ÿ/๐’™^(๐Ÿ ) )^(๐Ÿ‘/๐Ÿ) (๐ฅ๐จ๐ โก(๐Ÿ+๐Ÿ/๐’™^๐Ÿ )โˆ’๐Ÿ/๐Ÿ‘)+ C

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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.