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Ex 7.5, 11 - Integrate 5x / (x + 1) (x^2 - 4) - NCERT Maths

Ex 7.5, 11 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 11 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.5, 11 - Chapter 7 Class 12 Integrals - Part 4

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.