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

Ex 7.5, 23 - Chapter 7 Class 12 Integrals - Part 2
Ex 7.5, 23 - Chapter 7 Class 12 Integrals - Part 3 Ex 7.5, 23 - Chapter 7 Class 12 Integrals - Part 4

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.