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

Transcript

Misc 5 Integrate the function 1/( ๐‘ฅ^(1/2) + ๐‘ฅ^(1/3) ) โˆซ1โ–’1/(๐‘ฅ^(1/2) + ๐‘ฅ^(1/3) ) ๐‘‘๐‘ฅ Let x = ๐‘ก^6 ๐‘‘๐‘ฅ/๐‘‘๐‘ก=6๐‘ก^5 dx = 6๐‘ก^5 dt Substituting value of x and dx the equation โˆซ1โ–’1/(๐‘ฅ^(1/2) + ๐‘ฅ^(1/3) ) ๐‘‘๐‘ฅ = โˆซ1โ–’ใ€–6๐‘กใ€—^5/(ใ€–ใ€–(๐‘กใ€—^6) ใ€—^(1/2) +ใ€–ใ€– (๐‘กใ€—^6) ใ€—^(1/3) ) = โˆซ1โ–’ใ€–6๐‘กใ€—^5/(๐‘ก^3 + ๐‘ก^2 ) ๐‘‘๐‘ก = 6โˆซ1โ–’๐‘ก^3/(๐‘ก + 1) ๐‘‘๐‘ก Adding and subtracting 1 in numerator = 6โˆซ1โ–’(๐‘ก^3 + 1 โˆ’ 1)/(๐‘ก + 1) ๐‘‘๐‘ก = 6โˆซ1โ–’((๐‘ก^3 + 1 )/(๐‘ก + 1)โˆ’1/(๐‘ก + 1)) ๐‘‘๐‘ก Using ๐‘Ž^3+๐‘^3=(๐‘Ž+๐‘)(๐‘Ž^2+๐‘^2โˆ’๐‘Ž๐‘) = 6โˆซ1โ–’((๐‘ก + 1)(๐‘ก^2+1^2 โˆ’ 1 ร— ๐‘ก))/(๐‘ก + 1)โˆ’1/(๐‘ก + 1) ๐‘‘๐‘ก = 6โˆซ1โ–’ใ€–(๐‘ก^2+1โˆ’๐‘ก)โˆ’1/(1 + ๐‘ก)ใ€— ๐‘‘๐‘ก = 6[๐‘ก^3/3+๐‘กโˆ’๐‘ก^2/2โˆ’๐‘™๐‘œ๐‘”|1+๐‘ก|]+ C = 2๐‘ก^3+6๐‘กโˆ’ใ€–3๐‘กใ€—^2โˆ’6 ๐‘™๐‘œ๐‘”|1+๐‘ก|+ C Putting back value of t = 2 ใ€–ใ€–(๐‘ฅใ€—^(1/6))ใ€—^3+6ใ€–(๐‘ฅใ€—^(1/6))โˆ’3 ใ€–ใ€–(๐‘ฅใ€—^(1/6))ใ€—^2โˆ’6 ๐‘™๐‘œ๐‘”|1+๐‘ฅ^(1/6) |+ C = 2๐‘ฅ^(1/2)+6๐‘ฅ^(1/6)โˆ’3๐‘ฅ^(1/3)โˆ’6 ๐‘™๐‘œ๐‘”|1+๐‘ฅ^(1/6) |+ C = 2โˆš๐’™โˆ’๐Ÿ‘๐’™^(๐Ÿ/๐Ÿ‘)+ใ€–๐Ÿ” ๐’™ใ€—^(๐Ÿ/๐Ÿ”)โˆ’๐Ÿ” ๐ฅ๐จ๐ โกใ€–(๐Ÿ+๐’™^(๐Ÿ/๐Ÿ”))ใ€—+ C Because (1+ใ€–๐‘ฅ ใ€—^(1/6)) is always positive

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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.