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Ex 7.5, 7 - Integrate x / (x2 + 1) (x - 1) - Class 12 NCERT - Integration by partial fraction - Type 5

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise
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Ex 7.5, 7 π‘₯/(π‘₯^2 + 1)(π‘₯ βˆ’ 1) Let I=∫1β–’π‘₯/(π‘₯^2 + 1)(π‘₯ βˆ’ 1) 𝑑π‘₯ We can write integrand as π‘₯/(π‘₯^2 + 1)(π‘₯ βˆ’ 1) =(𝐴π‘₯ + 𝐡)/((π‘₯^2 + 1) ) + 𝐢/((π‘₯ βˆ’ 1) ) π‘₯/(π‘₯^2 + 1)(π‘₯ βˆ’ 1) =((𝐴π‘₯ + 𝐡)(π‘₯ βˆ’ 1) + 𝐢(π‘₯^2 + 1))/(π‘₯^2 + 1)(π‘₯ βˆ’ 1) By Cancelling denominator π‘₯=(𝐴π‘₯+𝐡)(π‘₯βˆ’1)+𝐢(π‘₯^2+1) Putting x = 1, in (1) π‘₯=(𝐴π‘₯+𝐡)(π‘₯βˆ’1)+𝐢(π‘₯^2+1) 1=(𝐴×1+𝐡)(1βˆ’1)+𝐢(1^2+1) 1=(𝐴+𝐡)Γ—0+𝐢(1+1) 1=0+2𝐢 𝐢=1/2 Putting x = 0 , in (1) π‘₯=(𝐴π‘₯+𝐡)(π‘₯βˆ’1)+𝐢(π‘₯^2+1) 0=(𝐴×0+𝐡)(0βˆ’1)+𝐢(0+1) 0 =𝐡(βˆ’1)+𝐢 ⇒𝐡=𝐢=1/2 Putting x = βˆ’1 βˆ’1=(𝐴(βˆ’1)+𝐡)(βˆ’1βˆ’1)+𝐢((1)^2+1) βˆ’1=(βˆ’π΄+𝐡)(βˆ’2)+𝐢(1+1) βˆ’1=2π΄βˆ’2𝐡+2𝐢 βˆ’1=2π΄βˆ’2Γ—1/2+2Γ—1/2 βˆ’1=2π΄βˆ’1+1 βˆ’1=2𝐴 𝐴=(βˆ’1)/2 Hence we can write π‘₯/(π‘₯^2 + 1)(π‘₯ βˆ’ 1) =((βˆ’ 1/2 π‘₯ + 1/2))/((π‘₯^2 + 1) ) + (1/2)/((π‘₯ βˆ’ 1) ) =(βˆ’1 . π‘₯)/(2 (π‘₯^2 + 1) ) + 1/(2 (π‘₯^2 + 1) )+1/2(π‘₯ βˆ’ 1) Integrating 𝑀.π‘Ÿ.𝑑.π‘₯ I=∫1β–’π‘₯/(π‘₯^2 + 1)(π‘₯ βˆ’ 1) 𝑑π‘₯ =βˆ’1/2 ∫1β–’π‘₯/(π‘₯^2 + 1) 𝑑π‘₯+1/2 ∫1▒𝑑π‘₯/(π‘₯^2 + 1) + 1/2 ∫1▒𝑑π‘₯/(π‘₯ βˆ’ 1) 𝑑π‘₯ Solving 𝐈𝟏 I1=βˆ’1/2 ∫1β–’π‘₯/((π‘₯^2 + 1) ) 𝑑π‘₯ Put π‘₯^2+1=𝑑 Different both sides w.r.t.π‘₯ 2π‘₯+0=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/2π‘₯ Hence we can write βˆ’1/2 ∫1β–’π‘₯/((π‘₯^2 + 1) ) 𝑑π‘₯=βˆ’1/2 ∫1β–’π‘₯/𝑑 𝑑𝑑/2π‘₯ =βˆ’1/(2 Γ— 2) ∫1▒𝑑𝑑/𝑑 =βˆ’1/4 log⁑|𝑑|+𝐢1 =βˆ’1/4 log⁑|π‘₯^2+1|+𝐢1 Solving 𝐈𝟐 I2=1/2 ∫1β–’1/(π‘₯^2 + 1) 𝑑π‘₯ =1/2 tan^(βˆ’1)⁑π‘₯+𝐢2 Solving πˆπŸ‘ I3=1/2 ∫1β–’1/(π‘₯ βˆ’ 1) 𝑑π‘₯ =1/2 log⁑|π‘₯βˆ’1|+𝐢3 Hence I=I1+I2+I3 =βˆ’1/4 log⁑|π‘₯^2+1|+𝐢1+1/2 tan^(βˆ’1)⁑π‘₯+𝐢2+1/2 log⁑|π‘₯βˆ’1|+𝐢3 =βˆ’1/4 log⁑|π‘₯^2+1|+1/2 tan^(βˆ’1)⁑π‘₯+1/2 log⁑|π‘₯βˆ’1|+𝐢 =βˆ’1/4 log⁑|π‘₯^2+1|+1/2 tan^(βˆ’1)⁑π‘₯+1/2 log⁑|π‘₯βˆ’1|+𝐢 =𝟏/𝟐 π’π’π’ˆβ‘|π’™βˆ’πŸ|βˆ’πŸ/πŸ’ γ€–π’π’π’ˆ 〗⁑|𝒙^𝟐+𝟏|+𝟏/𝟐 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑𝒙+ C

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