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  1. Class 12
  2. Important Question for exams Class 12

Transcript

Ex 7.5, 9 (3π‘₯ + 5)/(π‘₯^3 βˆ’ π‘₯^2 βˆ’ π‘₯ + 1) Let I=∫1β–’(3π‘₯ + 5)/(π‘₯^3 βˆ’ π‘₯^2 βˆ’ π‘₯ + 1) 𝑑π‘₯ We can write integrand as (3π‘₯ + 5)/(π‘₯^3 βˆ’ π‘₯^2 βˆ’ π‘₯ + 1)=(3π‘₯ + 5)/(π‘₯ βˆ’ 1)(π‘₯^2 βˆ’ 1) =(3π‘₯ + 5)/(π‘₯ βˆ’ 1)(π‘₯^2 βˆ’ 1^2 ) =(3π‘₯ + 5)/((π‘₯ βˆ’ 1) (π‘₯ βˆ’ 1) (π‘₯ + 1) ) =(3π‘₯ + 5)/γ€–(π‘₯ + 1) (π‘₯ βˆ’ 1)γ€—^2 Rough π‘₯^3βˆ’π‘₯^2βˆ’π‘₯+1 Put π‘₯=1 1^3βˆ’1^2βˆ’1+1 =1βˆ’1βˆ’1+1 =0 So, π‘₯βˆ’1 is a factor of π‘₯^3βˆ’π‘₯^2βˆ’π‘₯+1 We can write it as (3π‘₯ + 5)/γ€–(π‘₯ + 1) (π‘₯ βˆ’ 1)γ€—^2 =𝐴/((π‘₯ + 1) ) + 𝐡/((π‘₯ βˆ’ 1) ) + 𝐢/(π‘₯ βˆ’ 1)^2 (3π‘₯ + 5)/γ€–(π‘₯ + 1) (π‘₯ βˆ’ 1)γ€—^2 =(𝐴(π‘₯ βˆ’ 1)^2 + 𝐡(π‘₯ + 1)(π‘₯ βˆ’ 1) + 𝐢(π‘₯ + 1))/(π‘₯ + 1)(π‘₯ βˆ’ 1)(π‘₯ βˆ’ 1) (3π‘₯ + 5)/γ€–(π‘₯ + 1) (π‘₯ βˆ’ 1)γ€—^2 =(𝐴(π‘₯ βˆ’ 1)^2 + 𝐡(π‘₯^2 βˆ’ 1) + 𝐢(π‘₯ + 1))/((π‘₯ + 1) (π‘₯ βˆ’ 1)^2 ) By Cancelling denominator 3π‘₯+5=𝐴(π‘₯βˆ’1)^2+𝐡(π‘₯^2βˆ’1)+𝐢(π‘₯+1) Put π‘₯=1 , in (1) 3Γ—1+5=𝐴(1βˆ’1)^2+𝐡(1^2βˆ’1)+𝐢(1+1) 8=𝐴×0+ 𝐡×0+𝐢×2 8=2𝐢 𝐢=4 Similarly put π‘₯=βˆ’1 , in (1) 3π‘₯+5=𝐴(π‘₯βˆ’1)^2+𝐡(π‘₯^2βˆ’1)+𝐢(π‘₯+1) 3(βˆ’1)+5=𝐴(βˆ’1βˆ’1)^2+𝐡((βˆ’1)^2βˆ’1)+𝐢(βˆ’1+1) βˆ’3+5=𝐴(βˆ’2)^2+𝐡(1βˆ’1)+𝐢(0) 2=4𝐴+𝐡×0+𝐢×0 2=4𝐴 𝐴=1/2 Putting x = 0 , in (1) 3(0)+5=𝐴(0βˆ’1)^2+𝐡(0βˆ’1)+𝐢(0+1) 5=π΄βˆ’π΅+𝐢 5= 1/2 βˆ’π΅+4 5= βˆ’π΅+9/2 𝐡=9/2 βˆ’5 𝐡=βˆ’1/2 Hence we can write (3π‘₯ + 5)/γ€–(π‘₯ + 1) (π‘₯ βˆ’ 1)γ€—^2 =((1/2))/((π‘₯ + 1) ) + (βˆ’ 1/2)/((π‘₯ βˆ’ 1) ) + 4/(π‘₯ βˆ’ 1)^2 (3π‘₯ + 5)/γ€–(π‘₯ + 1) (π‘₯ βˆ’ 1)γ€—^2 =1/2(π‘₯ + 1) βˆ’ 1/2(π‘₯ βˆ’ 1) + 4/(π‘₯ βˆ’ 1)^2 Integrating 𝑀.π‘Ÿ.𝑑.π‘₯ I=∫1β–’(3π‘₯ + 5)/(π‘₯^3 βˆ’ π‘₯^2 βˆ’ π‘₯ + 1) 𝑑π‘₯=∫1β–’(1/2(π‘₯ + 1) βˆ’ 1/2(π‘₯ βˆ’ 1) + 4/(π‘₯ βˆ’ 1)^2 ) 𝑑π‘₯ =1/2 ∫1▒𝑑π‘₯/(π‘₯ + 1) βˆ’ 1/2 ∫1▒𝑑π‘₯/((π‘₯ βˆ’ 1) )+4∫1▒𝑑π‘₯/(π‘₯ βˆ’ 1)^2 Hence I=I1βˆ’I2+I3 Now, I1=1/2 ∫1β–’1/(π‘₯ + 1) 𝑑π‘₯ =1/2 log⁑|π‘₯+1|+𝐢1 Also, I2 =1/2 ∫1β–’1/(π‘₯ βˆ’ 1) 𝑑π‘₯ = 1/2 log⁑|π‘₯βˆ’1|+𝐢2 And, I3=∫1β–’4/(π‘₯ βˆ’ 1)^2 𝑑π‘₯ =4∫1β–’1/(π‘₯ βˆ’ 1)^2 𝑑π‘₯ =4∫1β–’(π‘₯ βˆ’ 1)^(βˆ’2) 𝑑π‘₯ =(4(π‘₯ βˆ’ 1)^(βˆ’2 + 1))/(βˆ’2 + 1) +𝐢3 =(4(π‘₯ βˆ’ 1)^(βˆ’1))/(βˆ’1) +𝐢3 =(βˆ’ 4)/(π‘₯ βˆ’ 1)+𝐢3 Therefore I=I1βˆ’I2+I3 I=1/2 log⁑|π‘₯+1|+𝐢1βˆ’ 1/2 log⁑|π‘₯βˆ’1|βˆ’πΆ2+(βˆ’ 4)/(π‘₯ βˆ’ 1)+𝐢3 =1/2 log⁑|π‘₯+1|+βˆ’ 1/2 log⁑|π‘₯βˆ’1|βˆ’4/(π‘₯ βˆ’ 1) +𝐢1βˆ’πΆ2+𝐢3 =1/2 [log⁑|π‘₯+1|βˆ’log⁑|π‘₯βˆ’1| ]βˆ’4/(π‘₯ βˆ’ 1) +𝐢 =𝟏/𝟐 π’π’π’ˆβ‘|(𝒙 + 𝟏)/(π’™βˆ’πŸ)|βˆ’ πŸ’/(𝒙 βˆ’ 𝟏) +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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.