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

Example 10 Find the following integrals: (i) ∫1β–’(π‘₯ + 2)/(2π‘₯^2 + 6π‘₯ + 5 ) 𝑑π‘₯ We can write numerator as π‘₯+2= A 𝑑/𝑑π‘₯ (2π‘₯^2+6π‘₯+5) + B π‘₯+2= A [4π‘₯+6]+ B π‘₯+2=4𝐴π‘₯+6A+B Finding A & B Comparing coefficient of π‘₯ π‘₯=4𝐴π‘₯ 1 =4A A=1/4 Comparing constant term 2=6A+B 2=6(1/4)+B 2=3/2+B B=2βˆ’3/2=1/2 Now, we know that π‘₯+2= A [4π‘₯+6]+ B π‘₯+2=1/4 [4π‘₯+6]+1/2 Now, our equation is ∫1β–’γ€–(π‘₯ + 2)/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯=∫1β–’γ€–(1/4 [4π‘₯ + 6] + 1/2)/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯γ€—γ€— =∫1β–’γ€–(1/4 [4π‘₯ + 6])/(2π‘₯^2 + 6π‘₯ + 5)+∫1β–’γ€–(1/2)/(2π‘₯^2+6π‘₯+5).𝑑π‘₯γ€—γ€— =1/4 ∫1β–’γ€–(4π‘₯ + 6)/(2π‘₯^2 + 6π‘₯ + 5) 𝑑π‘₯+1/2 ∫1β–’γ€–1/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯γ€—γ€— Solving I1 I1 =1/4 ∫1β–’γ€–(4π‘₯ + 6)/(2π‘₯^2 + 6π‘₯ + 5) 𝑑π‘₯γ€— Let t = 2π‘₯^2 + 6π‘₯ + 5 Differentiating both sides w.r.t.π‘₯ 4π‘₯ +6=𝑑𝑑/𝑑π‘₯ 𝑑π‘₯=𝑑𝑑/(4π‘₯ + 6) Now, I1 =1/4 ∫1β–’γ€–(4π‘₯ + 6)/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯γ€— Putting the values of (2π‘₯^2+6π‘₯+5) and 𝑑π‘₯, we get I1 =1/4 ∫1β–’γ€–(4π‘₯ + 6)/𝑑.𝑑𝑑/(4π‘₯ + 6) γ€— I1 =1/4 ∫1β–’γ€–1/𝑑.𝑑𝑑 γ€— I1 =1/4 π‘™π‘œπ‘”|𝑑|+𝐢1 I1 =1/4 π‘™π‘œπ‘”|2π‘₯^2+6π‘₯+5|+𝐢1 Solving I2 I2 =1/2 ∫1β–’γ€–1/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯ γ€— I2 =1/2 ∫1β–’γ€–1/2[π‘₯^2 + 6π‘₯/2 + 5/2 ] .𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/(π‘₯^2 +3π‘₯ + 5/2).𝑑π‘₯ γ€— (Using ∫1β–’γ€–1/π‘₯.𝑑π‘₯=π‘™π‘œπ‘”|π‘₯|+𝐢1γ€—) (Using 𝑑=2π‘₯^2+6π‘₯+5) I2 =1/4 ∫1β–’γ€–1/(π‘₯^2 + 2(π‘₯)(3/2) + 5/2).𝑑π‘₯ γ€— Adding & subtracting (3/2)^2 in denominator I2 =1/4 ∫1β–’γ€–1/(π‘₯^2 + 2(π‘₯) γ€–(3/2) +(3/2)^2βˆ’ (3/2)γ€—^2 + 5/2).𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2 βˆ’ (3/2)^2 + 5/2) 𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2 βˆ’ 9/4 + 5/2).𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2+ (βˆ’9 + 10)/4 ).𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2+ 1/4 ).𝑑π‘₯ γ€— I2 =1/4 ∫1β–’γ€–1/((π‘₯ + 3/2)^2+ (1/2)^2 ).𝑑π‘₯ γ€— =1/4 [1/(1/2) tan^(βˆ’1)⁑〖(π‘₯ + 3/2)/(1/2)+𝐢2γ€— ] =1/4 [2 tan^(βˆ’1)⁑〖((2π‘₯ + 3)/2)/(1/2)+𝐢2γ€— ] =1/4 [2 γ€–tan^(βˆ’1) (2π‘₯+3)〗⁑〖+𝐢2γ€— ] It is of form ∫1▒〖𝑑π‘₯/(π‘₯^2 + π‘Ž^2 )=1/π‘Ž γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑〖π‘₯/π‘Ž+𝐢2" " γ€— γ€— Replacing π‘₯ by (π‘₯+3/2) and by 1/2, we get =2/4 tan^(βˆ’1)⁑〖(2π‘₯+3)+𝐢2/4γ€— =1/2 tan^(βˆ’1)⁑〖(2π‘₯+3)+𝐢3γ€— Now, putting the value of I1 and I2 in eq. (1) ∴ ∫1β–’γ€–(π‘₯+2)/(2π‘₯^2 + 6π‘₯ + 5).𝑑π‘₯γ€— =1/4 π‘™π‘œπ‘”|2π‘₯^2+6π‘₯+5|+𝐢1+1/2 tan^(βˆ’1)⁑〖(2π‘₯+3)+γ€— 𝐢3 =𝟏/πŸ’ π’π’π’ˆ|πŸπ’™^𝟐+πŸ”π’™+πŸ“|+𝟏/𝟐 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑〖(πŸπ’™+πŸ‘)+γ€— π‘ͺ (where 𝐢3 = 𝐢2/4 )

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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.