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

Transcript

Ex 7.2, 7 Integrate the function: π‘₯√(π‘₯+2) π‘₯√(π‘₯+2) Step 1: Let (π‘₯+2)=𝑑 Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 1+0 = 𝑑𝑑/𝑑π‘₯ 1= 𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑 Step 2: Integrating the function ∫1β–’γ€–" " π‘₯√(π‘₯+2)γ€— .𝑑π‘₯ Putting the value of π‘₯+2 & 𝑑π‘₯ . = ∫1β–’γ€–π‘₯βˆšπ‘‘γ€— .𝑑π‘₯ = ∫1β–’γ€–π‘₯βˆšπ‘‘γ€— .𝑑𝑑 = ∫1β–’γ€–(π‘‘βˆ’2) βˆšπ‘‘γ€— .𝑑𝑑 = ∫1β–’γ€–(π‘‘βˆ’2) 𝑑^(1/2) γ€— .𝑑𝑑 = ∫1β–’(𝑑.𝑑^(1/2)βˆ’2.𝑑^(1/2) ) .𝑑𝑑 = ∫1β–’(𝑑^(3/2)βˆ’2.𝑑^(1/2) ) .𝑑𝑑 = ∫1▒𝑑^(3/2) .𝑑𝑑 βˆ’ 2∫1▒𝑑^(1/2) .𝑑𝑑 (Using π‘₯+2=𝑑, π‘₯=π‘‘βˆ’2) = 𝑑^(3/2 + 1)/(3/2 + 1) βˆ’ 2 . 𝑑^(1/2 + 1)/(1/2 + 1) + 𝐢 = 𝑑^(5/2)/(5/2) βˆ’ 2 . 𝑑^(3/2)/(3/2) + 𝐢 = 2/5 𝑑^(5/2) βˆ’ 2 Γ— 2/3 𝑑^(3/2) + 𝐢 = 2/5 𝑑^(5/2) βˆ’ 4/3 𝑑^(3/2) + 𝐢 Putting back 𝑑=π‘₯+2 = 𝟐/πŸ“ (𝒙+𝟐)^(πŸ“/𝟐) βˆ’ πŸ’/πŸ‘ (𝒙+𝟐)^(πŸ‘/𝟐) + π‘ͺ (Using ∫1β–’π‘₯^𝑛 . 𝑑π‘₯=π‘₯^(𝑛+1)/(𝑛 +1) )

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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.