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

Transcript

Ex 7.5, 11 Integrate the function 5π‘₯/((π‘₯ + 1) (π‘₯2βˆ’ 4) ) We can write the integrand as 5π‘₯/((π‘₯ + 1) (π‘₯2βˆ’ 4) ) = 5π‘₯/((π‘₯ + 1) (π‘₯ βˆ’ 2) (π‘₯ + 2) ) 5π‘₯/((π‘₯ + 1) (π‘₯2βˆ’ 4) ) = 𝐴/((π‘₯ + 1) ) + 𝐡/((π‘₯ βˆ’ 2) ) + 𝐢/((π‘₯ + 2) ) 5π‘₯/((π‘₯ + 1) (π‘₯2βˆ’ 4) ) = (𝐴(π‘₯ βˆ’ 2)(π‘₯ + 2) + 𝐡(π‘₯ + 1)(π‘₯ + 2) + 𝐢(π‘₯ +1)(π‘₯ βˆ’ 2))/((π‘₯ + 1) (π‘₯ βˆ’ 2) (π‘₯ + 2) ) Cancelling denominator 5π‘₯ = 𝐴(π‘₯βˆ’2)(π‘₯+2)+𝐡(π‘₯+1)(π‘₯+2)+𝐢(π‘₯+1)(π‘₯βˆ’2) …(1) Putting x = βˆ’1 in (1) 5π‘₯ = 𝐴(π‘₯βˆ’2)(π‘₯+2)+𝐡(π‘₯+1)(π‘₯+2)+𝐢(π‘₯+1)(π‘₯βˆ’2) 5( βˆ’1) = 𝐴(βˆ’1βˆ’2)(βˆ’1+2)+𝐡(βˆ’1+1)(βˆ’1+2)+𝐢(βˆ’1+1)(βˆ’1βˆ’2) βˆ’5 = 𝐴(βˆ’3)(1)+𝐡×0+𝐢×0 βˆ’5 = βˆ’3𝐴 𝐴 = (βˆ’5)/(βˆ’3) = 5/3 Putting x = 2 in (1) 5π‘₯ = 𝐴(π‘₯βˆ’2)(π‘₯+2)+𝐡(π‘₯+1)(π‘₯+2)+𝐢(π‘₯+1)(π‘₯βˆ’2) 5"(2) = " 𝐴(2βˆ’2)(2+2)+𝐡(2+1)(2+2)+𝐢(2+1)(2βˆ’2) 10 = 𝐴×0+𝐡(3)(4)+𝐢×0 10 = 12𝐡 𝐡 = 10/12=5/6 Putting x = βˆ’2 in (1) 5π‘₯ = 𝐴(π‘₯βˆ’2)(π‘₯+2)+𝐡(π‘₯+1)(π‘₯+2)+𝐢(π‘₯+1)(π‘₯βˆ’2) 5"("βˆ’"2) = " 𝐴(βˆ’2βˆ’2)(βˆ’2+2)+𝐡(βˆ’2+1)(βˆ’2+2)+𝐢(βˆ’2+1)(βˆ’2βˆ’2) βˆ’10 = 𝐴×0+𝐡×0+𝐢(βˆ’1)(βˆ’4) βˆ’10 = 4𝐢 𝐢 = (βˆ’10)/4 𝐢 = (βˆ’5)/2 Therefore ∫1β–’5π‘₯/((π‘₯ + 1) (π‘₯2βˆ’ 4) )=∫1β–’(𝐴/(π‘₯ + 1)+𝐡/(π‘₯ βˆ’ 2)+𝐢/(π‘₯ + 2)) 𝑑π‘₯ =5/3 ∫1▒𝑑π‘₯/(π‘₯ + 1) 𝑑π‘₯+ 5/6 ∫1▒𝑑π‘₯/(π‘₯ βˆ’ 2) 𝑑π‘₯βˆ’5/2 ∫1▒𝑑π‘₯/((π‘₯ + 2) ) =πŸ“/πŸ‘ γ€–π’π’π’ˆ 〗⁑|𝒙+𝟏|βˆ’ πŸ“/𝟐 γ€–π₯𝐨𝐠 〗⁑|𝒙+𝟐|+πŸ“/πŸ” γ€–π₯𝐨𝐠 〗⁑|π’™βˆ’π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.