Ex 7.5, 23 - Integrate dx / x (x^2 + 1) equals (A) log |x| - 1/2

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Ex 7.5, 23 - Chapter 7 Class 12 Integrals - Part 2

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Ex 7.5, 23 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.5, 23 - Chapter 7 Class 12 Integrals - Part 4

  1. Chapter 7 Class 12 Integrals (Term 2)
  2. Serial order wise

Transcript

Ex 7.5, 23 ∫1▒𝑑π‘₯/(π‘₯(π‘₯2 + 1) ) equals log |π‘₯| – 1/(2 ) log (x2+1) + C log |π‘₯| + 1/(2 ) log (x2+1) + C log |π‘₯| + 1/(2 ) log (x2+1) + C 1/(2 ) log |π‘₯| + log (x2+1) + C ∫1▒𝑑π‘₯/(π‘₯(π‘₯2 + 1) ) Let t = (1+π‘₯^2) Differentiating both sides 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑑/𝑑π‘₯ = 0+2π‘₯ 𝑑π‘₯ = 𝑑𝑑/2π‘₯ Thus, our equation becomes ∫1▒𝑑π‘₯/(π‘₯(π‘₯^2 + 1) ) = ∫1β–’1/(π‘₯ 𝑑) 𝑑𝑑/2π‘₯ = 1/2 ∫1β–’γ€– 𝑑𝑑/𝑑×1/π‘₯^2 γ€— = 1/2 ∫1▒𝑑𝑑/(𝑑(𝑑 βˆ’ 1)) We can write integrand as Since t = (1+π‘₯^2) x2 = (t – 1) 1/(𝑑(𝑑 βˆ’ 1)) = 𝐴/𝑑 + 𝐡/(𝑑 βˆ’ 1) 1/(𝑑(𝑑 βˆ’ 1)) = (𝐴(𝑑 βˆ’ 1) + 𝐡𝑑)/(𝑑(𝑑 βˆ’ 1)) Cancelling denominator 1 = 𝐴(π‘‘βˆ’1)+𝐡𝑑 Putting t = 0 in (1) 1 = 𝐴(0βˆ’1)+𝐡×0 1 = βˆ’π΄ 𝐴 = βˆ’1 Putting t = 1 in (1) 1 = 𝐴(1βˆ’1)+𝐡(1) 1 = 𝐴×0+𝐡 1 = 𝐡 𝐡 = 1 Therefore 1/2 ∫1▒𝑑𝑑/(𝑑(𝑑 βˆ’ 1)) = 1/2 ∫1β–’γ€–(βˆ’1)/𝑑 γ€— 𝑑𝑑 + 1/2 ∫1β–’γ€–1/(𝑑 βˆ’ 1) γ€— 𝑑𝑑 = (βˆ’1)/2 γ€–log 〗⁑|𝑑|+ 1/2 γ€–log 〗⁑|π‘‘βˆ’1|+𝐢 Putting back t = (1+π‘₯^2) = (βˆ’1)/2 γ€–log 〗⁑|1+π‘₯^2 |+ 1/2 γ€–log 〗⁑|1+π‘₯^2βˆ’1|+𝐢 = (βˆ’1)/2 γ€–log 〗⁑|1+π‘₯^2 |+ 1/2 γ€–log 〗⁑|π‘₯^2 |+𝐢 = (βˆ’1)/2 γ€–log 〗⁑|1+π‘₯^2 |+ 1/2Γ—2 γ€–log 〗⁑|π‘₯|+𝐢 = (βˆ’1)/2 γ€–log 〗⁑|1+π‘₯^2 |+ γ€–log 〗⁑|π‘₯|+𝐢 As 1 + π‘₯^2 it is always positive = γ€–log 〗⁑|π‘₯|βˆ’ 1/2 γ€–log 〗⁑〖(π‘₯^2+1)γ€—+𝐢 ∴ Correct answer is A . (log π‘₯^π‘Ž=π‘Ž log⁑π‘₯)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.