Example 40 - Evaluate [log (log x) + 1 / (log x)2 ] dx - Integration by parts

Slide26.JPG
Slide27.JPG Slide28.JPG

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

Transcript

Example 40 Evaluate ﷮﷮ log﷮ log﷮𝑥﷯﷯﷯+ 1﷮ log﷮𝑥﷯﷯﷮2﷯﷯﷯﷯𝑑𝑥 Let I1 = ﷮﷮ 𝑙𝑜𝑔 log﷮𝑥﷯﷯+ 1﷮ log﷮𝑥﷯﷯﷮2﷯﷯﷯𝑑𝑥﷯ I1 = ﷮﷮𝑙𝑜𝑔 log﷮𝑥﷯﷯𝑑𝑥+ ﷮﷮ 1﷮ log﷮𝑥﷯﷯﷮2﷯﷯.𝑑𝑥﷯﷯ Taking I2 i.e. I2 = ﷮﷮𝑙𝑜𝑔 log﷮𝑥﷯﷯𝑑𝑥﷯ I2 = ﷮﷮𝑙𝑜𝑔 log﷮𝑥﷯﷯.1 𝑑𝑥﷯ First Function, 𝑓 𝑥﷯=𝑙𝑜𝑔 𝑙𝑜𝑔𝑥﷯ Second Function, 𝑔 𝑥﷯=1 I2= ﷮﷮𝑙𝑜𝑔﷯ 𝑙𝑜𝑔𝑥﷯.1𝑑𝑥 I2=𝑙𝑜𝑔 𝑙𝑜𝑔𝑥﷯ ﷮﷮1.𝑑𝑥− ﷮﷮ 𝑑 𝑙𝑜𝑔 𝑙𝑜𝑔𝑥﷯﷯﷮𝑑𝑥﷯ ﷮﷮1.𝑑𝑥﷯﷯﷯﷯𝑑𝑥 I2=log 𝑙𝑜𝑔𝑥﷯ 𝑥﷯− ﷮﷮ 1﷮ log﷮𝑥﷯﷯. 𝑑 𝑙𝑜𝑔𝑥﷯﷮𝑑𝑥﷯ ﷮﷮1.𝑑𝑥﷯﷯𝑑𝑥﷯ I2=𝑥 𝑙𝑜𝑔 𝑙𝑜𝑔𝑥﷯− ﷮﷮ 1﷮ log﷮𝑥﷯﷯. 1﷮𝑥﷯ .𝑥 𝑑𝑥﷯ I2=𝑥 𝑙𝑜𝑔 𝑙𝑜𝑔𝑥﷯− ﷮﷮ 1﷮ log﷮𝑥﷯﷯. 1﷮𝑥﷯ .𝑥 𝑑𝑥﷯ I2=𝑥 𝑙𝑜𝑔 𝑙𝑜𝑔𝑥﷯− ﷮﷮ 1﷮ log﷮𝑥﷯﷯𝑑𝑥﷯ Taking I3 I3= ﷮﷮ 1﷮𝑙𝑜𝑔𝑥﷯.𝑑𝑥﷯ I3= ﷮﷮ 1﷮𝑙𝑜𝑔𝑥﷯.1 𝑑𝑥﷯ 𝑓 𝑥﷯= 1﷮𝑙𝑜𝑔𝑥﷯ & 𝑔 𝑥﷯=1 I3= 1﷮𝑙𝑜𝑔𝑥﷯ ﷮﷮1.𝑑𝑥− ﷮﷮ 𝑑 1﷮𝑙𝑜𝑔𝑥﷯﷯﷮𝑑𝑥﷯ ﷮﷮1.𝑑𝑥﷯﷯﷯﷯𝑑𝑥 I3= 1﷮𝑙𝑜𝑔𝑥﷯.𝑥− ﷮﷮ −1﷮ 𝑙𝑜𝑔𝑥﷯﷮2﷯﷯ 1﷮𝑥﷯﷯.𝑥 𝑑𝑥﷯ I3= 1﷮ log﷮𝑥﷯﷯+ ﷮﷮ 1﷮ log﷮𝑥﷯﷯﷮2﷯﷯ 𝑑𝑥﷯ Putting the value of I3 in I2 , we get I2=𝑥 𝑙𝑜𝑔 log﷮𝑥﷯﷯− 𝑥﷮ log﷮𝑥﷯﷯+ ﷮﷮ 1﷮ log﷮𝑥﷯﷯﷮2﷯﷯﷯𝑑𝑥﷯ I2=𝑥 𝑙𝑜𝑔 log﷮𝑥﷯﷯− 𝑥﷮ log﷮𝑥﷯﷯− ﷮﷮ 1﷮ log﷮𝑥﷯﷯﷮2﷯﷯𝑑𝑥﷯ Now, Putting the value of 𝐼2 in 𝐼1 , we get I1=𝑥 𝑙𝑜𝑔 log﷮𝑥﷯﷯− 𝑥﷮ log﷮𝑥﷯﷯− ﷮﷮ 1﷮ log﷮𝑥﷯﷯﷮2﷯﷯𝑑𝑥﷯+ ﷮﷮ 1﷮ log﷮𝑥﷯﷯﷮2﷯﷯𝑑𝑥﷯ ∴ I1=𝒙 𝒍𝒐𝒈 𝒍𝒐𝒈﷮𝒙﷯﷯− 𝒙﷮ 𝒍𝒐𝒈﷮𝒙﷯﷯ + C

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail