Example 38 - Chapter 7 Class 12 Integrals
Last updated at April 16, 2024 by Teachoo
Integration by parts
Ex 7.6, 3
Ex 7.6, 23 (MCQ)
Example 17
Ex 7.6, 1
Ex 7.6, 2 Important
Ex 7.6, 12
Example 21 Important
Ex 7.6, 21
Ex 7.6, 5 Important
Ex 7.6, 4
Ex 7.6, 6
Ex 7.6, 15
Example 18 Important
Ex 7.6, 14 Important
Ex 7.6, 7 Important
Ex 7.6, 9
Ex 7.6, 8
Ex 7.6, 11
Example 20 Important
Ex 7.6, 13 Important
Ex 7.6, 22 Important
Ex 7.6, 10 Important
Example 38 Important You are here
Integration by parts
Last updated at April 16, 2024 by Teachoo
Example 38 Evaluate ∫1▒[log(log𝑥 )+1/(log𝑥 )^2 ] 𝑑𝑥 Let I1 =∫1▒[𝑙𝑜𝑔(log𝑥 )+1/(log𝑥 )^2 ]𝑑𝑥 I1 = ∫1▒〖𝑙𝑜𝑔(log𝑥 )𝑑𝑥+∫1▒〖1/(log𝑥 )^2 .𝑑𝑥〗〗 Solving 𝐈𝟐 I2 =∫1▒𝑙𝑜𝑔(log𝑥 )𝑑𝑥 I2 =∫1▒〖𝑙𝑜𝑔(log𝑥 ).1 𝑑𝑥〗 Let I1 =∫1▒[𝑙𝑜𝑔(log𝑥 )+1/(log𝑥 )^2 ]𝑑𝑥 I1 = ∫1▒〖𝑙𝑜𝑔(log𝑥 )𝑑𝑥+∫1▒〖1/(log𝑥 )^2 .𝑑𝑥〗〗 Solving 𝐈𝟐 I2 =∫1▒𝑙𝑜𝑔(log𝑥 )𝑑𝑥 I2 =∫1▒〖𝑙𝑜𝑔(log𝑥 ).1 𝑑𝑥〗 I2=𝑙𝑜𝑔(𝑙𝑜𝑔𝑥) ∫1▒〖1.𝑑𝑥−∫1▒[𝑑[𝑙𝑜𝑔[𝑙𝑜𝑔𝑥]]/𝑑𝑥 ∫1▒〖1.𝑑𝑥〗] 〗 𝑑𝑥 I2=log(𝑙𝑜𝑔𝑥)(𝑥)−∫1▒[1/log𝑥 .𝑑(𝑙𝑜𝑔𝑥)/𝑑𝑥 ∫1▒〖1.𝑑𝑥〗]𝑑𝑥 I2=𝑥 𝑙𝑜𝑔(𝑙𝑜𝑔𝑥)−∫1▒〖1/log𝑥 .1/𝑥 .𝑥 𝑑𝑥〗 I2=𝑥 𝑙𝑜𝑔(𝑙𝑜𝑔𝑥)−∫1▒〖1/log𝑥 .1/𝑥 .𝑥 𝑑𝑥〗 I2=𝑥 𝑙𝑜𝑔(𝑙𝑜𝑔𝑥)−∫1▒〖1/log𝑥 𝑑𝑥〗 Solving 𝐈𝟑 I3=∫1▒〖1/log𝑥 .𝑑𝑥〗 I3=∫1▒〖1/log𝑥 .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/log𝑥 +∫1▒〖1/(log𝑥 )^2 𝑑𝑥〗 Putting the value of I3 in I2 , we get I2=𝑥 𝑙𝑜𝑔(log𝑥 )−𝐼3 I2=𝑥 𝑙𝑜𝑔(log𝑥 )−[𝑥/log𝑥 +∫1▒1/(log𝑥 )^2 𝑑𝑥] I2=𝑥 𝑙𝑜𝑔(log𝑥 )−𝑥/log𝑥 −∫1▒〖1/(log𝑥 )^2 𝑑𝑥〗 Now, Putting the value of 𝐼2 in 𝐼1 , we get I1=I2+∫1▒〖1/(log𝑥 )^2 𝑑𝑥〗 I1=𝑥 𝑙𝑜𝑔(log𝑥 )−𝑥/log𝑥 −∫1▒〖1/(log𝑥 )^2 𝑑𝑥〗+∫1▒〖1/(log𝑥 )^2 𝑑𝑥〗 ∴ I1=𝒙 𝒍𝒐𝒈(𝒍𝒐𝒈𝒙 )−𝒙/𝒍𝒐𝒈𝒙 + C