Chapter 7 Class 12 Integrals
Concept wise

    Slide26.JPG

Slide27.JPG
Slide28.JPG Slide29.JPG Slide30.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Example 38 Evaluate ∫1▒[log⁡(log⁡𝑥 )+1/(log⁡𝑥 )^2 ] 𝑑𝑥 Let I1 =∫1▒[𝑙𝑜𝑔(log⁡𝑥 )+1/(log⁡𝑥 )^2 ]𝑑𝑥 I1 = ∫1▒〖𝑙𝑜𝑔(log⁡𝑥 )𝑑𝑥+∫1▒〖1/(log⁡𝑥 )^2 .𝑑𝑥〗〗 Solving 𝐈𝟐 I2 =∫1▒𝑙𝑜𝑔(log⁡𝑥 )𝑑𝑥 I2 =∫1▒〖𝑙𝑜𝑔(log⁡𝑥 ).1 𝑑𝑥〗 Let I1 =∫1▒[𝑙𝑜𝑔(log⁡𝑥 )+1/(log⁡𝑥 )^2 ]𝑑𝑥 I1 = ∫1▒〖𝑙𝑜𝑔(log⁡𝑥 )𝑑𝑥+∫1▒〖1/(log⁡𝑥 )^2 .𝑑𝑥〗〗 Solving 𝐈𝟐 I2 =∫1▒𝑙𝑜𝑔(log⁡𝑥 )𝑑𝑥 I2 =∫1▒〖𝑙𝑜𝑔(log⁡𝑥 ).1 𝑑𝑥〗 I2=𝑙𝑜𝑔(𝑙𝑜𝑔𝑥) ∫1▒〖1.𝑑𝑥−∫1▒[𝑑[𝑙𝑜𝑔[𝑙𝑜𝑔𝑥]]/𝑑𝑥 ∫1▒〖1.𝑑𝑥〗] 〗 𝑑𝑥 I2=log(𝑙𝑜𝑔𝑥)(𝑥)−∫1▒[1/log⁡𝑥 .𝑑(𝑙𝑜𝑔𝑥)/𝑑𝑥 ∫1▒〖1.𝑑𝑥〗]𝑑𝑥 I2=𝑥 𝑙𝑜𝑔(𝑙𝑜𝑔𝑥)−∫1▒〖1/log⁡𝑥 .1/𝑥 .𝑥 𝑑𝑥〗 I2=𝑥 𝑙𝑜𝑔(𝑙𝑜𝑔𝑥)−∫1▒〖1/log⁡𝑥 .1/𝑥 .𝑥 𝑑𝑥〗 I2=𝑥 𝑙𝑜𝑔(𝑙𝑜𝑔𝑥)−∫1▒〖1/log⁡𝑥 𝑑𝑥〗 Solving 𝐈𝟑 I3=∫1▒〖1/log⁡𝑥 .𝑑𝑥〗 I3=∫1▒〖1/log⁡𝑥 .1 𝑑𝑥〗 I3=1/𝑙𝑜𝑔𝑥 ∫1▒〖1.𝑑𝑥−∫1▒[𝑑(1/𝑙𝑜𝑔𝑥)/𝑑𝑥 ∫1▒〖1.𝑑𝑥〗] 〗 𝑑𝑥 I3=1/𝑙𝑜𝑔𝑥.𝑥−∫1▒〖(−1)/(𝑙𝑜𝑔𝑥)^2 (1/𝑥).𝑥 𝑑𝑥〗 I3=𝑥/log⁡𝑥 +∫1▒〖1/(log⁡𝑥 )^2 𝑑𝑥〗 Putting the value of I3 in I2 , we get I2=𝑥 𝑙𝑜𝑔(log⁡𝑥 )−𝐼3 I2=𝑥 𝑙𝑜𝑔(log⁡𝑥 )−[𝑥/log⁡𝑥 +∫1▒1/(log⁡𝑥 )^2 𝑑𝑥] I2=𝑥 𝑙𝑜𝑔(log⁡𝑥 )−𝑥/log⁡𝑥 −∫1▒〖1/(log⁡𝑥 )^2 𝑑𝑥〗 Now, Putting the value of 𝐼2 in 𝐼1 , we get I1=I2+∫1▒〖1/(log⁡𝑥 )^2 𝑑𝑥〗 I1=𝑥 𝑙𝑜𝑔(log⁡𝑥 )−𝑥/log⁡𝑥 −∫1▒〖1/(log⁡𝑥 )^2 𝑑𝑥〗+∫1▒〖1/(log⁡𝑥 )^2 𝑑𝑥〗 ∴ I1=𝒙 𝒍𝒐𝒈(𝒍𝒐𝒈⁡𝒙 )−𝒙/𝒍𝒐𝒈⁡𝒙 + C

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.