Check sibling questions

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

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

Slide1.JPG

Slide2.JPG
Slide3.JPG

This video is only available for Teachoo black users

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


Transcript

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

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

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.