Chapter 7 Class 12 Integrals
Concept wise

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Misc 23 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 = 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€— = 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€— 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.