Check sibling questions

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Transcript

Misc 1 (Method 1) Integrate the function 1/(π‘₯ βˆ’ π‘₯^3 ) Solving integrand 1/(π‘₯ βˆ’ π‘₯^3 )=1/π‘₯(1 βˆ’ π‘₯^2 ) =𝟏/𝒙(𝟏 βˆ’ 𝒙)(𝟏 + 𝒙) We can write it as 𝟏/𝒙(𝟏 βˆ’ 𝒙)(𝟏 + 𝒙) = 𝑨/𝒙 + 𝑩/((𝟏 βˆ’ 𝒙) ) + 𝒄/((𝟏 + 𝒙) ) 1/π‘₯(1 βˆ’ π‘₯)(1 + π‘₯) = (𝐴(1 βˆ’ π‘₯) (1 + π‘₯) + 𝐡π‘₯ (1 + π‘₯) + 𝐢π‘₯ (1 βˆ’ π‘₯))/( π‘₯ (1 βˆ’ π‘₯) (1 + π‘₯) ) Cancelling denominator 𝟏 = 𝑨(𝟏 βˆ’ 𝒙) (𝟏 + 𝒙) + 𝑩𝒙 (𝟏 + 𝒙) + π‘ͺ𝒙 (𝟏 βˆ’ 𝒙) Putting 𝒙=𝟎 in (1) 1=𝐴(1 βˆ’ 0) (1 + 0) + 𝐡(0) (1 + 0) + 𝐢(0) (1 βˆ’ 0) 1=𝐴 Γ— 1 Γ— 1 + 𝐡 Γ— 0 + 𝐢 Γ— 0 1=𝐴+0+0 𝑨=𝟏 Putting 𝒙=𝟏 in (1) 1 = 𝐴(1 βˆ’1) (1 +1) + 𝐡(1) (1 +1) + 𝐢(1) (1 βˆ’1) 1 = 𝐴 Γ— 0 + 𝐡 Γ— (1) Γ— (2) + 𝐢 Γ— 0 1 = 0 +2𝐡 + 0 𝑩 = 𝟏/𝟐 Putting 𝒙=βˆ’πŸ in (1) 1 = 𝐴(1 βˆ’(βˆ’1)) (1 +(βˆ’1)) + 𝐡(βˆ’1) (1 +(βˆ’1)) + 𝐢(βˆ’1) (1 βˆ’(βˆ’1)) 1 = 𝐴(1 + 1) (1 βˆ’1) + 𝐡(βˆ’1)(1βˆ’1) + 𝐢(βˆ’1)(1+1) 1 = 𝐴 Γ— 0 + 𝐡 Γ— 0 + 𝐢 Γ—(βˆ’1)(2) 1 = 0 +0 βˆ’2𝐢 1 = βˆ’2𝐢 π‘ͺ = βˆ’πŸ/𝟐 Hence we can write it as 1/π‘₯(1 βˆ’ π‘₯)(1 + π‘₯) = 1/π‘₯ + (𝟏/𝟐)/((1 βˆ’ π‘₯) ) + (βˆ’1/2)/((1 + π‘₯) ) 𝟏/𝒙(𝟏 βˆ’ 𝒙)(𝟏 + 𝒙) = 𝟏/𝒙 + 𝟏/𝟐(𝟏 βˆ’ 𝒙) + (βˆ’πŸ)/𝟐(𝟏 + 𝒙) Therefore ∫1β–’πŸ/𝒙(𝟏 βˆ’ 𝒙)(𝟏 + 𝒙) 𝒅𝒙 = ∫1β–’1/π‘₯ 𝑑π‘₯ + ∫1β–’1/2(1 βˆ’ π‘₯) 𝑑π‘₯ + ∫1β–’(βˆ’1)/2(1 + π‘₯) 𝑑π‘₯ = ∫1β–’1/π‘₯ 𝑑π‘₯ + 1/2 ∫1β–’1/((1 βˆ’ π‘₯) ) 𝑑π‘₯ βˆ’ 1/2 ∫1β–’1/((1 + π‘₯) ) 𝑑π‘₯ = γ€–π₯𝐨𝐠 〗⁑|𝒙|+𝟏/𝟐 [γ€–π₯𝐨𝐠 〗⁑|𝟏 βˆ’ 𝒙|/(βˆ’πŸ)] βˆ’1/2 γ€–π₯𝐨𝐠 〗⁑|𝟏 + 𝒙|+π‘ͺ = γ€–log 〗⁑|π‘₯|βˆ’ γ€– 1/2 log 〗⁑|1 βˆ’ π‘₯|βˆ’1/2 γ€–log 〗⁑|1 + π‘₯|+𝐢 = γ€–log 〗⁑|π‘₯|βˆ’1/2 [γ€–log 〗⁑|1 βˆ’ π‘₯|+γ€–log 〗⁑|1 + π‘₯| ]+𝐢 = γ€–log 〗⁑|π‘₯|βˆ’1/2 [γ€–log 〗⁑|1 βˆ’ π‘₯| |1 + π‘₯|]+𝐢 As π’π’π’ˆ 𝑨+π’π’π’ˆ 𝑩=log⁑𝐴𝐡 = γ€–log 〗⁑|π‘₯|βˆ’1/2 [γ€–log 〗⁑|(1 βˆ’ π‘₯^2 )| ]+𝐢 = 1/2 [2 γ€–log 〗⁑|π‘₯|βˆ’γ€–log 〗⁑|(1 βˆ’ π‘₯^2 )|+2𝐢] = 1/2 [γ€–log 〗⁑〖|π‘₯|^2 γ€—βˆ’γ€–log 〗⁑|(1 βˆ’ π‘₯^2 )|+𝐾] = 𝟏/𝟐 log |𝒙^𝟐/((𝟏 βˆ’ 𝒙^𝟐 ) )|+𝑲 As π’π’π’ˆ π‘¨βˆ’π’π’π’ˆ 𝑩=π‘™π‘œπ‘” 𝐴/𝐡 Misc 1 (Method 2) Integrate the function 1/(π‘₯ βˆ’ π‘₯^3 ) Now, ∫1β–’γ€–1/(π‘₯ βˆ’ π‘₯^3 ) 𝑑π‘₯γ€— Taking x3 common from the denominator =∫1β–’1/(π‘₯^3 (π‘₯/π‘₯^3 βˆ’ 1) ) 𝑑π‘₯ =∫1β–’πŸ/(𝒙^πŸ‘ (𝟏/𝒙^𝟐 βˆ’ 𝟏) ) 𝒅𝒙 Let t = 𝟏/𝒙^𝟐 βˆ’πŸ Differentiating with respect to π‘₯ 𝑑/𝑑π‘₯ (1/π‘₯^2 βˆ’1)=𝑑𝑑/𝑑π‘₯ (βˆ’2)/π‘₯^3 =𝑑𝑑/𝑑π‘₯ 𝒅𝒙=(𝒙^πŸ‘ 𝒅𝒕)/(βˆ’πŸ) Putting the value t and dt in the equation ∫1β–’γ€–πŸ/(𝒙^πŸ‘ (𝟏/𝒙^𝟐 βˆ’πŸ) ) 𝒅𝒙〗=∫1β–’γ€–1/(π‘₯^3 (𝑑) ) Γ— (π‘₯^3 𝑑𝑑)/(βˆ’2)γ€— =∫1β–’γ€–πŸ/(βˆ’πŸ) 𝒅𝒕/𝒕〗 =(βˆ’1)/( 2) ∫1▒𝑑𝑑/𝑑 =(βˆ’1)/( 2) π‘™π‘œπ‘”|𝑑|+𝐢 Putting back 𝒕=𝟏/𝒙^𝟐 βˆ’πŸ =(βˆ’1)/( 2) π‘™π‘œπ‘”|1/π‘₯^2 βˆ’1|+𝐢 =(βˆ’1)/( 2) π‘™π‘œπ‘”|(𝟏 βˆ’ 𝒙^𝟐)/𝒙^𝟐 |+𝐢 = 1/2 log |(1 βˆ’ π‘₯^2)/π‘₯^2 |^(βˆ’πŸ)+𝐢 = 𝟏/𝟐 log |𝒙^𝟐/(𝟏 βˆ’ 𝒙^𝟐 )|+π‘ͺ (As a log b = log 𝑏^π‘Ž)

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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.