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Misc 24 - Integrate root x2+1 [ log (x2 + 1) - 2 log x / x4

Misc 24 - Chapter 7 Class 12 Integrals - Part 2
Misc 24 - Chapter 7 Class 12 Integrals - Part 3 Misc 24 - Chapter 7 Class 12 Integrals - Part 4 Misc 24 - Chapter 7 Class 12 Integrals - Part 5


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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.