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Find f log x (1 + log x) 2 dx

This question is similar to Ex 7.2, 35 - Chapter 7 Class 12 - Integrals

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Question 1 – Choice 1 Find ∫1β–’γ€–log⁑π‘₯/(1 + log⁑π‘₯ )^2 𝑑π‘₯γ€—Let 𝐈=∫1β–’γ€–log⁑π‘₯/(1 + log⁑π‘₯ )^2 𝑑π‘₯γ€— =∫1β–’γ€–(π’π’π’ˆβ‘π’™ + 𝟏 βˆ’ 𝟏)/(1 + log⁑π‘₯ )^2 𝑑π‘₯γ€— =∫1β–’γ€–((1 + log⁑π‘₯) βˆ’ 1)/(1 + log⁑π‘₯ )^2 𝑑π‘₯γ€— =∫1β–’γ€–((1 + log⁑π‘₯) )/(1 + log⁑π‘₯ )^2 𝑑π‘₯γ€—βˆ’βˆ«1β–’γ€–1/(1 + log⁑π‘₯ )^2 𝑑π‘₯γ€— =∫1β–’γ€–(𝟏 )/((𝟏 + π’π’π’ˆβ‘π’™ ) ) π’…π’™γ€—βˆ’βˆ«1β–’γ€–πŸ/(𝟏 + π’π’π’ˆβ‘π’™ )^𝟐 𝒅𝒙〗 Solving ∫1β–’γ€–(𝟏 )/((𝟏 + π₯𝐨𝐠⁑𝐱 ) ) 𝐝𝐱〗 Using Integration by parts ∫1β–’γ€–(𝟏 )/((𝟏 + π’π’π’ˆβ‘π’™ ) ) 𝒅𝒙〗 = ∫1β–’γ€–(1 )/((1 + π‘™π‘œπ‘”β‘π‘₯ ) ) Γ— 1 𝑑π‘₯γ€— = 1/((1 + log⁑π‘₯)) ∫1β–’γ€–1 𝑑π‘₯γ€—βˆ’βˆ«1β–’(𝒅(𝟏/(𝟏 + π’π’π’ˆ 𝒙))/𝒅𝒙 ∫1β–’γ€–πŸ 𝒅𝒙〗) 𝒅𝒙 = 1/((1 + log⁑π‘₯))Γ— π‘₯βˆ’βˆ«1β–’((βˆ’πŸ)/(𝟏 +π’π’π’ˆ 𝒙)^𝟐 Γ—πŸ/𝒙 Γ— 𝒙) 𝒅𝒙 = π‘₯/((1 + log⁑π‘₯))+∫1β–’πŸ/(𝟏 + π’π’π’ˆ 𝒙)^𝟐 𝒅𝒙We know that ∫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/(1 + log x) and g(x) = 1 Thus I=∫1β–’γ€–(1 )/((1 + π‘™π‘œπ‘”β‘π‘₯ ) ) 𝑑π‘₯γ€—βˆ’βˆ«1β–’γ€–1/(1 + π‘™π‘œπ‘”β‘π‘₯ )^2 𝑑π‘₯γ€— = π‘₯/((1 + log⁑π‘₯))+∫1β–’1/(1 + π‘™π‘œπ‘” π‘₯)^2 𝑑π‘₯βˆ’βˆ«1β–’γ€–1/(1 + π‘™π‘œπ‘”β‘π‘₯ )^2 𝑑π‘₯γ€— = 𝒙/((𝟏 + π’π’π’ˆβ‘π’™))+π‘ͺ

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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.