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

Transcript

Example 40 Evaluate โˆซ1โ–’[logโก(logโก๐‘ฅ )+1/(logโก๐‘ฅ )^2 ] ๐‘‘๐‘ฅ Let I1 =โˆซ1โ–’[๐‘™๐‘œ๐‘”(logโก๐‘ฅ )+1/(logโก๐‘ฅ )^2 ]๐‘‘๐‘ฅ I1 = โˆซ1โ–’ใ€–๐‘™๐‘œ๐‘”(logโก๐‘ฅ )๐‘‘๐‘ฅ+โˆซ1โ–’ใ€–1/(logโก๐‘ฅ )^2 .๐‘‘๐‘ฅใ€—ใ€— Solving ๐ˆ๐Ÿ I2 =โˆซ1โ–’๐‘™๐‘œ๐‘”(logโก๐‘ฅ )๐‘‘๐‘ฅ I2 =โˆซ1โ–’ใ€–๐‘™๐‘œ๐‘”(logโก๐‘ฅ ).1 ๐‘‘๐‘ฅใ€— Using by parts โˆซ1โ–’ใ€–๐‘“(๐‘ฅ) ๐‘”โก(๐‘ฅ) ใ€— ๐‘‘๐‘ฅ=๐‘“(๐‘ฅ) โˆซ1โ–’๐‘”(๐‘ฅ) ๐‘‘๐‘ฅโˆ’โˆซ1โ–’(๐‘“^โ€ฒ (๐‘ฅ) โˆซ1โ–’๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ) ๐‘‘๐‘ฅ Putting f(x) = log (log x) and g(x) = 1 I2=๐‘™๐‘œ๐‘”(๐‘™๐‘œ๐‘”๐‘ฅ) โˆซ1โ–’ใ€–1.๐‘‘๐‘ฅโˆ’โˆซ1โ–’[๐‘‘[๐‘™๐‘œ๐‘”[๐‘™๐‘œ๐‘”๐‘ฅ]]/๐‘‘๐‘ฅ โˆซ1โ–’ใ€–1.๐‘‘๐‘ฅใ€—] ใ€— ๐‘‘๐‘ฅ I2=log(๐‘™๐‘œ๐‘”๐‘ฅ)(๐‘ฅ)โˆ’โˆซ1โ–’[1/logโก๐‘ฅ .๐‘‘(๐‘™๐‘œ๐‘”๐‘ฅ)/๐‘‘๐‘ฅ โˆซ1โ–’ใ€–1.๐‘‘๐‘ฅใ€—]๐‘‘๐‘ฅ I2=๐‘ฅ ๐‘™๐‘œ๐‘”(๐‘™๐‘œ๐‘”๐‘ฅ)โˆ’โˆซ1โ–’ใ€–1/logโก๐‘ฅ .1/๐‘ฅ .๐‘ฅ ๐‘‘๐‘ฅใ€— I2=๐‘ฅ ๐‘™๐‘œ๐‘”(๐‘™๐‘œ๐‘”๐‘ฅ)โˆ’โˆซ1โ–’ใ€–1/logโก๐‘ฅ .1/๐‘ฅ .๐‘ฅ ๐‘‘๐‘ฅใ€— I2=๐‘ฅ ๐‘™๐‘œ๐‘”(๐‘™๐‘œ๐‘”๐‘ฅ)โˆ’โˆซ1โ–’ใ€–1/logโก๐‘ฅ ๐‘‘๐‘ฅใ€— Solving ๐ˆ๐Ÿ‘ I3=โˆซ1โ–’ใ€–1/logโก๐‘ฅ .๐‘‘๐‘ฅใ€— I3=โˆซ1โ–’ใ€–1/logโก๐‘ฅ .1 ๐‘‘๐‘ฅใ€— I3=1/๐‘™๐‘œ๐‘”๐‘ฅ โˆซ1โ–’ใ€–1.๐‘‘๐‘ฅโˆ’โˆซ1โ–’[๐‘‘(1/๐‘™๐‘œ๐‘”๐‘ฅ)/๐‘‘๐‘ฅ โˆซ1โ–’ใ€–1.๐‘‘๐‘ฅใ€—] ใ€— ๐‘‘๐‘ฅ I3=1/๐‘™๐‘œ๐‘”๐‘ฅ.๐‘ฅโˆ’โˆซ1โ–’ใ€–(โˆ’1)/(๐‘™๐‘œ๐‘”๐‘ฅ)^2 (1/๐‘ฅ).๐‘ฅ ๐‘‘๐‘ฅใ€— Using by parts โˆซ1โ–’ใ€–๐‘“(๐‘ฅ) ๐‘”โก(๐‘ฅ) ใ€— ๐‘‘๐‘ฅ=๐‘“(๐‘ฅ) โˆซ1โ–’๐‘”(๐‘ฅ) ๐‘‘๐‘ฅโˆ’โˆซ1โ–’(๐‘“^โ€ฒ (๐‘ฅ) โˆซ1โ–’๐‘”(๐‘ฅ) ๐‘‘๐‘ฅ) ๐‘‘๐‘ฅ Putting f(x) = 1/๐‘™๐‘œ๐‘”โก๐‘ฅ and g(x) = 1 I3=๐‘ฅ/logโก๐‘ฅ +โˆซ1โ–’ใ€–1/(logโก๐‘ฅ )^2 ๐‘‘๐‘ฅใ€— Putting the value of I3 in I2 , we get I2=๐‘ฅ ๐‘™๐‘œ๐‘”(logโก๐‘ฅ )โˆ’๐ผ3 I2=๐‘ฅ ๐‘™๐‘œ๐‘”(logโก๐‘ฅ )โˆ’[๐‘ฅ/logโก๐‘ฅ +โˆซ1โ–’1/(logโก๐‘ฅ )^2 ๐‘‘๐‘ฅ] I2=๐‘ฅ ๐‘™๐‘œ๐‘”(logโก๐‘ฅ )โˆ’๐‘ฅ/logโก๐‘ฅ โˆ’โˆซ1โ–’ใ€–1/(logโก๐‘ฅ )^2 ๐‘‘๐‘ฅใ€— Now, Putting the value of ๐ผ2 in ๐ผ1 , we get I1=I2+โˆซ1โ–’ใ€–1/(logโก๐‘ฅ )^2 ๐‘‘๐‘ฅใ€— I1=๐‘ฅ ๐‘™๐‘œ๐‘”(logโก๐‘ฅ )โˆ’๐‘ฅ/logโก๐‘ฅ โˆ’โˆซ1โ–’ใ€–1/(logโก๐‘ฅ )^2 ๐‘‘๐‘ฅใ€—+โˆซ1โ–’ใ€–1/(logโก๐‘ฅ )^2 ๐‘‘๐‘ฅใ€— โˆด I1=๐’™ ๐’๐’๐’ˆ(๐’๐’๐’ˆโก๐’™ )โˆ’๐’™/๐’๐’๐’ˆโก๐’™ + 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.