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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Misc 3 Integrate the function 1/(π‘₯ √(π‘Žπ‘₯ βˆ’ π‘₯^2 ) ) ∫1β–’1/(π‘₯ √(π‘Žπ‘₯ βˆ’ π‘₯^2 ) ) 𝑑π‘₯ = ∫1β–’1/(π‘₯ √(π‘₯^2 (π‘Ž/π‘₯ βˆ’ 1) ) ) 𝑑π‘₯ = ∫1β–’1/(π‘₯ . π‘₯√((π‘Ž/π‘₯ βˆ’ 1) ) ) 𝑑π‘₯ = ∫1β–’1/(π‘₯^2 √((π‘Ž/π‘₯ βˆ’ 1) ) ) 𝑑π‘₯ Let π‘Ž/π‘₯ βˆ’1=𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ (βˆ’ π‘Ž)/π‘₯^2 βˆ’0 = 𝑑𝑑/𝑑π‘₯ (βˆ’ π‘Ž)/π‘₯^2 = 𝑑𝑑/𝑑π‘₯ 𝑑π‘₯ = (βˆ’ π‘₯^2)/π‘Ž . 𝑑𝑑 Putting the values of (a/xβˆ’1) and dx, we get ∫1β–’1/(π‘₯^2 √((π‘Ž/π‘₯ βˆ’ 1) ) ) 𝑑π‘₯ = ∫1β–’1/(π‘₯^2 βˆšπ‘‘ ) 𝑑π‘₯ = ∫1β–’1/(π‘₯^2 βˆšπ‘‘ )Γ—(βˆ’π‘₯^2)/π‘Ž . 𝑑𝑑 = ∫1β–’(βˆ’1)/π‘Ž 𝑑𝑑/βˆšπ‘‘ = (βˆ’1)/π‘Ž ∫1β–’(𝑑)^(βˆ’ 1/2) 𝑑𝑑 = (βˆ’1)/π‘Ž [𝑑^((βˆ’1)/2 + 1)/((βˆ’1)/2 + 1)]+𝐢 = (βˆ’1)/π‘Ž [𝑑^(1/2)/(1/2)] + 𝐢 = (βˆ’2)/π‘Ž [βˆšπ‘‘] + 𝐢 Putting back t = π‘Ž/π‘₯βˆ’1 = (βˆ’2)/π‘Ž √(π‘Ž/π‘₯ βˆ’1) + 𝐢 = (βˆ’πŸ)/𝒂 √((𝒂 βˆ’ 𝒙)/𝒙) + π‘ͺ

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.