Integration Full Chapter Explained - https://you.tube/Integration-Class-12

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  1. Chapter 7 Class 12 Integrals
  2. Serial order wise

Transcript

Misc 1 Integrate the function 1/(π‘₯ βˆ’ π‘₯^3 ) We can it write as 1/(π‘₯ βˆ’ π‘₯^3 )=(1/π‘₯^3 )/(((π‘₯ βˆ’ π‘₯^3 ))/π‘₯^3 ) =1/(π‘₯^3 (π‘₯/π‘₯^3 βˆ’ π‘₯^3/π‘₯^3 ) ) =1/(π‘₯^3 (1/π‘₯^2 βˆ’1) ) Dividing by π‘₯^3 both numerator and denominator Integrating w.r.t. π‘₯ ∫1β–’γ€–1/(π‘₯ βˆ’ π‘₯^3 ) 𝑑π‘₯γ€—=∫1β–’1/(π‘₯^3 (1/π‘₯^2 βˆ’ 1) ) 𝑑π‘₯ Let t = 1/π‘₯^2 βˆ’1 Differentiating with respect to π‘₯ 𝑑/𝑑π‘₯ (1/π‘₯^2 βˆ’1)=𝑑𝑑/𝑑π‘₯ (βˆ’2)/π‘₯^3 =𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=(π‘₯^3 𝑑𝑑)/(βˆ’2) Putting the value t and dt in the equation ∫1β–’γ€–1/(π‘₯^3 (1/π‘₯^2 βˆ’1) ) 𝑑π‘₯γ€—=∫1β–’γ€–1/(π‘₯^3 (𝑑) ) Γ— (π‘₯^3 𝑑𝑑)/(βˆ’2)γ€— =∫1β–’γ€–1/(βˆ’2) 𝑑𝑑/𝑑〗 =(βˆ’1)/( 2) ∫1▒𝑑𝑑/𝑑 =(βˆ’1)/( 2) π‘™π‘œπ‘”|𝑑|+𝐢 Putting back 𝑑=1/π‘₯^2 βˆ’1 =(βˆ’1)/( 2) π‘™π‘œπ‘”|1/π‘₯^2 βˆ’1|+𝐢 =(βˆ’1)/( 2) π‘™π‘œπ‘”|1/π‘₯^2 βˆ’π‘₯^2/π‘₯^2 |+𝐢 =(βˆ’1)/( 2) π‘™π‘œπ‘”|(1 βˆ’ π‘₯^2)/π‘₯^2 |+𝐢 = 1/2 log |(1 βˆ’ π‘₯^2)/π‘₯^2 |^(βˆ’1)+𝐢 = 𝟏/𝟐 log |𝒙^𝟐/(𝟏 βˆ’ 𝒙^𝟐 )|+π‘ͺ We know that 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 9 years. He provides courses for Maths and Science at Teachoo.