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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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.