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Misc 5 - Integrate 1 / x1/2 + x1/3 - Class 12 NCERT - Miscellaneous

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


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 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.