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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.