Check sibling questions

Example 40 - Evaluate [log (log x) + 1 / (log x)2 ] dx - Examples

Example 40 - Chapter 7 Class 12 Integrals - Part 2
Example 40 - Chapter 7 Class 12 Integrals - Part 3 Example 40 - Chapter 7 Class 12 Integrals - Part 4 Example 40 - Chapter 7 Class 12 Integrals - Part 5


Transcript

Example 40 Evaluate ∫1β–’[log⁑(log⁑π‘₯ )+1/(log⁑π‘₯ )^2 ] 𝑑π‘₯ Let I1 =∫1β–’[π‘™π‘œπ‘”(log⁑π‘₯ )+1/(log⁑π‘₯ )^2 ]𝑑π‘₯ I1 = ∫1β–’γ€–π‘™π‘œπ‘”(log⁑π‘₯ )𝑑π‘₯+∫1β–’γ€–1/(log⁑π‘₯ )^2 .𝑑π‘₯γ€—γ€— Solving 𝐈𝟐 I2 =∫1β–’π‘™π‘œπ‘”(log⁑π‘₯ )𝑑π‘₯ I2 =∫1β–’γ€–π‘™π‘œπ‘”(log⁑π‘₯ ).1 𝑑π‘₯γ€— Using by parts ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓^β€² (π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = log (log x) and g(x) = 1 I2=π‘™π‘œπ‘”(π‘™π‘œπ‘”π‘₯) ∫1β–’γ€–1.𝑑π‘₯βˆ’βˆ«1β–’[𝑑[π‘™π‘œπ‘”[π‘™π‘œπ‘”π‘₯]]/𝑑π‘₯ ∫1β–’γ€–1.𝑑π‘₯γ€—] γ€— 𝑑π‘₯ I2=log(π‘™π‘œπ‘”π‘₯)(π‘₯)βˆ’βˆ«1β–’[1/log⁑π‘₯ .𝑑(π‘™π‘œπ‘”π‘₯)/𝑑π‘₯ ∫1β–’γ€–1.𝑑π‘₯γ€—]𝑑π‘₯ I2=π‘₯ π‘™π‘œπ‘”(π‘™π‘œπ‘”π‘₯)βˆ’βˆ«1β–’γ€–1/log⁑π‘₯ .1/π‘₯ .π‘₯ 𝑑π‘₯γ€— I2=π‘₯ π‘™π‘œπ‘”(π‘™π‘œπ‘”π‘₯)βˆ’βˆ«1β–’γ€–1/log⁑π‘₯ .1/π‘₯ .π‘₯ 𝑑π‘₯γ€— I2=π‘₯ π‘™π‘œπ‘”(π‘™π‘œπ‘”π‘₯)βˆ’βˆ«1β–’γ€–1/log⁑π‘₯ 𝑑π‘₯γ€— Solving πˆπŸ‘ I3=∫1β–’γ€–1/log⁑π‘₯ .𝑑π‘₯γ€— I3=∫1β–’γ€–1/log⁑π‘₯ .1 𝑑π‘₯γ€— I3=1/π‘™π‘œπ‘”π‘₯ ∫1β–’γ€–1.𝑑π‘₯βˆ’βˆ«1β–’[𝑑(1/π‘™π‘œπ‘”π‘₯)/𝑑π‘₯ ∫1β–’γ€–1.𝑑π‘₯γ€—] γ€— 𝑑π‘₯ I3=1/π‘™π‘œπ‘”π‘₯.π‘₯βˆ’βˆ«1β–’γ€–(βˆ’1)/(π‘™π‘œπ‘”π‘₯)^2 (1/π‘₯).π‘₯ 𝑑π‘₯γ€— Using by parts ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓^β€² (π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = 1/π‘™π‘œπ‘”β‘π‘₯ and g(x) = 1 I3=π‘₯/log⁑π‘₯ +∫1β–’γ€–1/(log⁑π‘₯ )^2 𝑑π‘₯γ€— Putting the value of I3 in I2 , we get I2=π‘₯ π‘™π‘œπ‘”(log⁑π‘₯ )βˆ’πΌ3 I2=π‘₯ π‘™π‘œπ‘”(log⁑π‘₯ )βˆ’[π‘₯/log⁑π‘₯ +∫1β–’1/(log⁑π‘₯ )^2 𝑑π‘₯] I2=π‘₯ π‘™π‘œπ‘”(log⁑π‘₯ )βˆ’π‘₯/log⁑π‘₯ βˆ’βˆ«1β–’γ€–1/(log⁑π‘₯ )^2 𝑑π‘₯γ€— Now, Putting the value of 𝐼2 in 𝐼1 , we get I1=I2+∫1β–’γ€–1/(log⁑π‘₯ )^2 𝑑π‘₯γ€— I1=π‘₯ π‘™π‘œπ‘”(log⁑π‘₯ )βˆ’π‘₯/log⁑π‘₯ βˆ’βˆ«1β–’γ€–1/(log⁑π‘₯ )^2 𝑑π‘₯γ€—+∫1β–’γ€–1/(log⁑π‘₯ )^2 𝑑π‘₯γ€— ∴ I1=𝒙 π’π’π’ˆ(π’π’π’ˆβ‘π’™ )βˆ’π’™/π’π’π’ˆβ‘π’™ + C

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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.