Ex 7.5, 9 - Integrate 3x + 5 / x^3 - x^2 - x + 1 - Teachoo

Ex 7.5, 9 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 9 - Chapter 7 Class 12 Integrals - Part 3
Ex 7.5, 9 - Chapter 7 Class 12 Integrals - Part 4
Ex 7.5, 9 - Chapter 7 Class 12 Integrals - Part 5 Ex 7.5, 9 - Chapter 7 Class 12 Integrals - Part 6 Ex 7.5, 9 - Chapter 7 Class 12 Integrals - Part 7 Ex 7.5, 9 - Chapter 7 Class 12 Integrals - Part 8


Transcript

Ex 7.5, 9 Integrate the function (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 …(1) 8=2𝐢 𝐢=4 Putting 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) 3π‘₯+5=𝐴(π‘₯βˆ’1)^2+𝐡(π‘₯^2βˆ’1)+𝐢(π‘₯+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 our equation as write (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 =((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)/(𝒙 βˆ’ 𝟏)|βˆ’ πŸ’/(𝒙 βˆ’ 𝟏) +π‘ͺ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.