Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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

Transcript

Ex 7.4, 22 Integrate the function (π‘₯ + 3)/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) ∫1β–’(π‘₯ + 3)/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) 𝑑π‘₯ =1/2 ∫1β–’(2π‘₯ + 6)/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) 𝑑π‘₯ =1/2 ∫1β–’(2π‘₯ βˆ’ 2 + 2 + 6 )/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) 𝑑π‘₯ =1/2 ∫1β–’(2π‘₯ βˆ’ 2)/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) 𝑑π‘₯+8/2 ∫1▒𝑑π‘₯/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) =1/2 ∫1β–’(2π‘₯ βˆ’ 2)/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) 𝑑π‘₯+4∫1▒𝑑π‘₯/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) Rough (π‘₯^2βˆ’2π‘₯βˆ’5)^β€²=2π‘₯βˆ’2 Solving π‘°πŸ I1=1/2 ∫1β–’(2π‘₯ βˆ’ 2)/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) . 𝑑π‘₯ Let π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5=𝑑 Diff both sides w.r.t.x 2π‘₯βˆ’2βˆ’0=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(2π‘₯ βˆ’ 2) Thus, our equation becomes ∴ I1=1/2 ∫1β–’(2π‘₯ βˆ’ 2)/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) . 𝑑π‘₯ Putting value of (π‘₯^2βˆ’2π‘₯βˆ’5)=𝑑 and 𝑑π‘₯=𝑑𝑑/(2π‘₯ βˆ’ 2) I1=1/2 ∫1β–’(2π‘₯ βˆ’ 2)/𝑑 . 𝑑π‘₯ I1=1/2 ∫1β–’(2π‘₯ βˆ’ 2)/𝑑 . 𝑑𝑑/(2π‘₯ βˆ’ 2) I1=1/2 ∫1β–’1/𝑑 . 𝑑𝑑 I1=1/2 log⁑|𝑑|+𝐢1 I1=1/2 log⁑|π‘₯^2βˆ’2π‘₯βˆ’5|+𝐢1 Solving π‘°πŸ I2=4∫1β–’1/(π‘₯^2 βˆ’ 2π‘₯ βˆ’ 5) . 𝑑π‘₯ (Using 𝑑=π‘₯^2βˆ’2π‘₯βˆ’5) I2=4∫1β–’1/(π‘₯^2 βˆ’ 2(π‘₯)(1) βˆ’ 5) . 𝑑π‘₯ I2=4∫1β–’1/(π‘₯^2 βˆ’ 2(π‘₯)(1) + (1)^2 βˆ’ (1)^2 βˆ’ 5) . 𝑑π‘₯ I2=4∫1β–’1/((π‘₯ βˆ’ 1)^2 βˆ’ (1)^2 βˆ’ 5) . I2=4∫1β–’1/((π‘₯ βˆ’ 1)^2 βˆ’ 1 βˆ’ 5) . 𝑑π‘₯ I2=4∫1β–’1/((π‘₯ βˆ’ 1)^2 βˆ’ 6) . 𝑑π‘₯ I2=4∫1β–’1/((π‘₯ βˆ’ 1)^2 βˆ’(√6 )^2 ) . 𝑑π‘₯ It is of form ∫1▒𝑑π‘₯/(π‘₯^2 βˆ’ π‘Ž^2 ) =1/2π‘Ž log⁑|(π‘₯ βˆ’ π‘Ž)/(π‘₯ + π‘Ž)|+𝐢 ∴ Replacing π‘₯ by (π‘₯βˆ’1) and a by √6 , we get I2=4/(2√6) log⁑|(π‘₯ βˆ’ 1 βˆ’ √6)/(π‘₯ βˆ’ 1 + √6)|+𝐢2 I2=2/√6 log⁑|(π‘₯ βˆ’ 1 βˆ’ √6)/(π‘₯ βˆ’ 1 + √6)|+𝐢2 Putting the values of I1 and I2 in (1) ∫1β–’γ€–(π‘₯ + 2)/√(π‘₯^2 + 2π‘₯ + 3).γ€— . 𝑑π‘₯ = 𝐼_1+𝐼_2 =1/2 log⁑|π‘₯^2βˆ’2π‘₯βˆ’5|+𝐢1+2/√6 log⁑|(π‘₯ βˆ’ 1 βˆ’ √6)/(π‘₯ βˆ’ 1 + √6)|+𝐢"2 " =𝟏/𝟐 π’π’π’ˆβ‘|𝒙^πŸβˆ’πŸπ’™βˆ’πŸ“|+𝟐/βˆšπŸ” π’π’π’ˆβ‘|(𝒙 βˆ’ 𝟏 βˆ’ βˆšπŸ”)/(𝒙 βˆ’ 𝟏 + βˆšπŸ”)|+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.