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Find ∫1▒γ€–(x + 1)/((x^2  + 1)  x) dxγ€—

This question is similar to Ex 13.2, 2 - Chapter 13 Class 12 - Probability

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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 7 Find ∫1β–’γ€–(π‘₯ + 1)/((π‘₯^2 + 1) π‘₯) 𝑑π‘₯γ€—Let (π‘₯ + 1)/((π‘₯^2 + 1) π‘₯) = (𝑨𝒙 + 𝑩)/((𝒙^𝟐 + 𝟏) ) + π‘ͺ/𝒙 (π‘₯ + 1)/((π‘₯^2 + 1) π‘₯) = ((𝐴π‘₯ + 𝐡)π‘₯ + 𝐢(1 + π‘₯^2 ))/((π‘₯^2 + 1) π‘₯) By cancelling denominator 𝒙 + 𝟏 = (𝑨𝒙 + 𝑩)𝒙 + π‘ͺ(𝟏 + 𝒙^𝟐 ) Putting 𝒙=𝟎 0 + 1 = (𝐴(0) + 𝐡) Γ— 0 + 𝐢(1 +0^2 ) 1 = 𝐢 π‘ͺ = 𝟏 Putting 𝒙=𝟏 1 + 1 = (𝐴(1) + 𝐡) Γ— 1 + 𝐢(1 +1^2 ) 2 = (𝐴 + 𝐡) +2𝐢 Putting 𝐢 = 1 2 = (𝐴 + 𝐡) +2 Γ— 1 2 = (𝐴 + 𝐡) +2 2βˆ’2 = (𝐴 + 𝐡) 0 = 𝐴 + 𝐡 𝑨=βˆ’ 𝑩 Putting 𝒙=βˆ’πŸ βˆ’1 + 1 = (𝐴(βˆ’1) + 𝐡) Γ— βˆ’1 + 𝐢(1 +γ€–(βˆ’1)γ€—^2 ) 0 = βˆ’(βˆ’π΄ + 𝐡) +𝐢 Γ— (1+1) 0 = βˆ’(βˆ’π΄ + 𝐡) +2𝐢 Putting 𝐴=βˆ’ 𝐡 0 = βˆ’(𝐡 + 𝐡) +2𝐢 0 = βˆ’2B +2𝐢 2B =2𝐢 B =𝐢 Putting 𝐢 = 1 𝑩 = 𝟏 And, 𝐴=βˆ’π΅ ∴ 𝑨=βˆ’πŸ Thus, 𝐴=βˆ’1, 𝐡=1, 𝐢 = 1 So, we can write (𝒙 + 𝟏)/((𝒙^𝟐 + 𝟏) 𝒙) = (𝐴π‘₯ + 𝐡)/((π‘₯^2 + 1) ) + 𝐢/π‘₯ = ((βˆ’1)π‘₯ +1)/((π‘₯^2 + 1) ) + 1/π‘₯ = (βˆ’π’™ + 𝟏)/((𝒙^𝟐 + 𝟏) ) + 𝟏/𝒙 Therefore integrating ∫1β–’(π‘₯ + 1)/((π‘₯^2 + 1) π‘₯) 𝑑π‘₯ = ∫1β–’(βˆ’π’™ + 𝟏)/((𝒙^𝟐 + 𝟏) ) 𝑑π‘₯ + ∫1β–’1/(π‘₯ ) 𝑑π‘₯ = ∫1β–’(βˆ’π‘₯ + 1)/((π‘₯^2 + 1) ) 𝑑π‘₯ + ∫1β–’1/(π‘₯ ) 𝑑π‘₯ = ∫1β–’(βˆ’π‘₯)/((π‘₯^2 + 1) ) 𝑑π‘₯ + ∫1β–’1/(π‘₯^2 + 1) 𝑑π‘₯ + ∫1β–’1/(π‘₯ ) 𝑑π‘₯ = ∫1β–’(βˆ’π‘₯)/((π‘₯^2 + 1) ) 𝑑π‘₯ + γ€–π­πšπ§γ€—^(βˆ’πŸ)⁑𝒙 + π’π’π’ˆβ‘γ€–|𝒙|γ€—+𝐢 = ∫1β–’(βˆ’π‘₯)/((π‘₯^2 + 1) ) 𝑑π‘₯ + γ€–π­πšπ§γ€—^(βˆ’πŸ)⁑𝒙 + π’π’π’ˆβ‘γ€–|𝒙|γ€—+𝐢 Solving 𝐈1 I1 = ∫1β–’(βˆ’π‘₯)/(π‘₯^2 + 1) 𝑑π‘₯ Let 𝒕 = 𝒙^𝟐+𝟏 𝑑𝑑/𝑑π‘₯ = 2π‘₯ 𝑑𝑑/2π‘₯ = 𝑑π‘₯ Hence ∫1β–’(βˆ’π‘₯)/(π‘₯^2 + 1) 𝑑π‘₯ = ∫1β–’γ€–(βˆ’π‘₯)/𝑑 . 𝑑𝑑/2π‘₯γ€— = βˆ’βˆ«1▒𝑑𝑑/2(𝑑) = (βˆ’1)/2 γ€–log 〗⁑|𝑑|+𝐢1 Putting back t = π‘₯^2+1 = (βˆ’πŸ)/𝟐 γ€–π’π’π’ˆ 〗⁑|𝒙^𝟐+𝟏|+π‘ͺ𝟐 Therefore integrating ∫1β–’(π‘₯ + 1)/((π‘₯^2 + 1) π‘₯) 𝑑π‘₯ = (βˆ’πŸ)/𝟐 γ€–π’π’π’ˆ 〗⁑|𝒙^𝟐+𝟏| + γ€–π­πšπ§γ€—^(βˆ’πŸ)⁑𝒙 + π’π’π’ˆβ‘γ€–|𝒙|γ€—+𝐢

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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.