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Chapter 7 Class 12 Integrals
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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

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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

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.