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

Transcript

Example 2 Find the following integrals: (i) βˆ«β–’(π‘₯^3 βˆ’ 1)/π‘₯^2 𝑑π‘₯ ∫1β–’γ€–((π‘₯^3 βˆ’ 1))/π‘₯^2 .𝑑π‘₯γ€— =∫1β–’(π‘₯^3/π‘₯^2 βˆ’1/π‘₯^2 )𝑑π‘₯ =∫1β–’γ€–(π‘₯βˆ’1/π‘₯^2 ).𝑑π‘₯γ€— =∫1β–’γ€–π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’γ€–1/π‘₯^2 𝑑π‘₯γ€—γ€— =∫1β–’γ€–π‘₯^1 𝑑π‘₯βˆ’βˆ«1β–’γ€–π‘₯^(βˆ’2). 𝑑π‘₯γ€—γ€— =[(π‘₯^(1 + 1)/(1 + 1))]βˆ’[(π‘₯^(βˆ’2 + 1)/(βˆ’2 + 1))]+𝐢 =π‘₯^2/2βˆ’π‘₯^(βˆ’1)/(βˆ’1)+𝐢 (β–ˆ(∫1β–’γ€–π‘₯^𝑛 𝑑π‘₯=π‘₯^(𝑛+1)/(𝑛+1)+𝐢〗)) =π‘₯^2/2+π‘₯^(βˆ’1)+𝐢 =𝒙^𝟐/𝟐+𝟏/𝒙 +π‘ͺ Example 2 Find the following integrals: (ii) βˆ«β–’(π‘₯^(2/3)+1) 𝑑π‘₯ ∫1β–’(π‘₯^(2/3)+1) 𝑑π‘₯ =∫1β–’(π‘₯^(2/3)+π‘₯^0 )𝑑π‘₯ =∫1β–’π‘₯^(2/3) 𝑑π‘₯+∫1β–’π‘₯^0 𝑑π‘₯ = π‘₯^(2/3 + 1)/(2/3 + 1)+ π‘₯^(0 + 1)/(0 + 1)+𝐢 = π‘₯^(5/3)/(5/3)+π‘₯+𝐢 = γ€–πŸ‘π’™γ€—^(πŸ“/πŸ‘)/πŸ“+𝒙+π‘ͺ ("As" ∫1β–’γ€–π‘₯^𝑛 𝑑π‘₯=π‘₯^(𝑛+1)/(𝑛+1)+𝐢〗) Example 2 Find the following integrals: (iii) βˆ«β–’(π‘₯^(3/2)+2𝑒^π‘₯βˆ’1/π‘₯) 𝑑π‘₯ ∫1β–’(π‘₯^(3/2)+2𝑒^π‘₯βˆ’1/π‘₯) 𝑑π‘₯ =∫1β–’π‘₯^(3/2) 𝑑π‘₯+2∫1▒𝑒^π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’1/π‘₯ 𝑑π‘₯ = π‘₯^(3/2 + 1)/(3/2 + 1) +2𝑒^π‘₯βˆ’log⁑〖 |π‘₯|γ€—+𝐢 = π‘₯^(5/2)/(5/2)+2𝑒^π‘₯βˆ’γ€–log 〗⁑|π‘₯|+𝐢 = γ€–πŸπ’™ γ€—^(πŸ“/𝟐)/πŸ“+πŸπ’†^π’™βˆ’γ€–π’π’π’ˆ 〗⁑|𝒙|+π‘ͺ As ∫1β–’π‘₯^𝑛 𝑑π‘₯=π‘₯^(𝑛 + 1)/(𝑛 + 1) +𝐢 ∫1▒𝑒^π‘₯ 𝑑π‘₯=𝑒^π‘₯ +𝐢 ∫1β–’1/π‘₯ 𝑑π‘₯=log⁑〖 |π‘₯|γ€— +𝐢

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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.