Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Ex 7.5 (2024)

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Ex 7.5

Ex 7.5, 1

Ex 7.5, 2

Ex 7.5, 3 Important

Ex 7.5, 4

Ex 7.5, 5

Ex 7.5, 6 Important You are here

Ex 7.5, 7 Important

Ex 7.5, 8

Ex 7.5, 9 Important

Ex 7.5, 10

Ex 7.5, 11 Important

Ex 7.5, 12

Ex 7.5, 13 Important

Ex 7.5, 14 Important

Ex 7.5, 15

Ex 7.5, 16 Important

Ex 7.5, 17

Ex 7.5, 18 Important

Ex 7.5, 19

Ex 7.5, 20 Important

Ex 7.5, 21 Important

Ex 7.5, 22 (MCQ)

Ex 7.5, 23 (MCQ) Important

Ex 7.6β†’

Chapter 7 Class 12 Integrals

Serial order wise

  • Ex 7.1
  • Ex 7.2
  • Ex 7.3
  • Ex 7.4
  • Ex 7.5

  • Ex 7.6
  • Ex 7.7
  • Ex 7.8
  • Ex 7.9
  • Ex 7.10
  • Examples
  • Miscellaneous
  • Case Based Questions (MCQ)
  • NCERT Exemplar MCQ
  • Area as a sum

Last updated at April 16, 2024 by Teachoo

Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Ex 7.5 (2)

Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Ex 7.5 (3)

Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Ex 7.5 (4)

Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Ex 7.5 (5)

Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Ex 7.5 (6)

Transcript

Ex 7.5, 6Integrate the function (1 βˆ’ π‘₯2)/(π‘₯(1 βˆ’ 2π‘₯)) ∫1β–’(1 βˆ’ π‘₯2)/(π‘₯(1 βˆ’ 2π‘₯)) 𝑑π‘₯= ∫1β–’(1 βˆ’ π‘₯^2)/(π‘₯ βˆ’ 2π‘₯^2 ) 𝑑π‘₯ =∫1β–’(1/2 + (βˆ’ π‘₯/2 + 1)/(βˆ’2π‘₯^2+ π‘₯)) 𝑑π‘₯ =1/2 ∫1▒𝑑π‘₯+1/2 ∫1β–’(βˆ’π‘₯ + 2)/(βˆ’2π‘₯^2+ π‘₯) 𝑑π‘₯ =1/2 ∫1▒𝑑π‘₯βˆ’1/2 ∫1β–’(π‘₯ βˆ’ 2)/π‘₯(1 βˆ’ 2π‘₯) 𝑑π‘₯ =1/2 ∫1▒𝑑π‘₯βˆ’1/2 ∫1β–’(π‘₯ βˆ’ 2)/π‘₯(2π‘₯ βˆ’ 1) 𝑑π‘₯ βˆ’2π‘₯^2+π‘₯ βˆ’π‘₯^2+π‘₯/2Now Solving (π‘₯ βˆ’ 2)/(π‘₯ (2π‘₯ βˆ’ 1) ) = 𝐴/π‘₯ + 𝐡/(2π‘₯ βˆ’ 1) (π‘₯ βˆ’ 2)/(π‘₯ (2π‘₯ βˆ’ 1) ) = (𝐴(2π‘₯ βˆ’ 1) + 𝐡π‘₯)/(π‘₯ (2π‘₯ βˆ’ 1) )Cancelling denominator π‘₯βˆ’2=𝐴(2π‘₯βˆ’1)+𝐡π‘₯Putting x = 0 in (2) π‘₯βˆ’2=𝐴(2π‘₯βˆ’1)+𝐡π‘₯0βˆ’2 = 𝐴(2Γ—0βˆ’1) + 𝐡×0 βˆ’2 = A(βˆ’1)…(2) βˆ’2 = βˆ’π΄ 𝐴 = 2Similarly Putting x = 1/2 in (2) π‘₯βˆ’2=𝐴(2π‘₯βˆ’1)+𝐡π‘₯ 1/2 βˆ’ 2 = A(1/2Γ—2βˆ’1)+𝐡×1/2 (βˆ’3)/2 = A(1βˆ’1)+𝐡×1/2 (βˆ’3)/2 = AΓ—0+ 𝐡/2 (βˆ’3)/2 = 𝐡/2𝐡 = βˆ’3Hence we can write it as (π‘₯ βˆ’ 2)/(π‘₯ (2π‘₯ βˆ’ 1) ) = 𝐴/π‘₯ + 𝐡/(2π‘₯ βˆ’ 1)(π‘₯ βˆ’ 2)/(π‘₯ (2π‘₯ βˆ’ 1) ) = 2/π‘₯ + ((βˆ’3))/(2π‘₯ βˆ’ 1) = 2/π‘₯ βˆ’ 3/(2π‘₯ βˆ’ 1)Therefore , from (1) we get, ∫1β–’(1 βˆ’ π‘₯^2)/π‘₯(1 βˆ’ 2π‘₯) =1/2 ∫1▒𝑑π‘₯+ 1/2 ∫1β–’(2/π‘₯ βˆ’ 3/(2π‘₯ βˆ’ 1)) 𝑑π‘₯ =1/2 ∫1▒𝑑π‘₯+ 2/2 ∫1▒〖𝑑π‘₯/π‘₯ βˆ’3/2γ€— ∫1▒𝑑π‘₯/(2π‘₯ βˆ’ 1) =1/2 ∫1▒𝑑π‘₯+∫1▒〖𝑑π‘₯/π‘₯ + 3/2γ€— ∫1▒𝑑π‘₯/(1 βˆ’ 2π‘₯) =1/2 π‘₯+log⁑|π‘₯|+3/2 log⁑|1 βˆ’ 2π‘₯|/(βˆ’2) +𝐢 =𝒙/𝟐 +π’π’π’ˆβ‘|𝒙|βˆ’πŸ‘/πŸ’ π’π’π’ˆβ‘|πŸβˆ’πŸπ’™|+π‘ͺ

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Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Ex 7.5 (7)

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Ex 7.5, 6 - Integrate 1 - x2 / x (1 - 2x) - Class 12 - Ex 7.5 (2024)

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