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

Transcript

Example 39 Evaluate ∫1β–’(π‘₯^4 𝑑π‘₯)/(π‘₯ βˆ’1)(π‘₯^2 + 1) Let I = ∫1β–’(π‘₯^4 𝑑π‘₯)/(π‘₯ βˆ’1)(π‘₯^2 + 1) 𝑑π‘₯ We can write π‘₯^4/(π‘₯ βˆ’1)(π‘₯^2 + 1) = π‘₯^4/(π‘₯^3 βˆ’ π‘₯^2+ π‘₯ βˆ’ 1) Dividing Numerator by denominator as follows. Hence π‘₯^4 = (π‘₯^3βˆ’π‘₯^2+π‘₯+1) (π‘₯+1)+1 Thus, π‘₯^4/(π‘₯^3 βˆ’ π‘₯^2 + π‘₯ + 1) = (π‘₯+1)+1/(π‘₯^3 βˆ’ π‘₯^2 + π‘₯ + 1) = (π‘₯+1)+1/((π‘₯ βˆ’ 1) (π‘₯^2 +1) ) Now, we can write 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) )= (𝐴π‘₯ + 𝐡)/(π‘₯^2 + 1) + 𝐢/(π‘₯ βˆ’ 1) 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) )= ((𝐴π‘₯ + 𝐡)(π‘₯ βˆ’ 1) + 𝐢 (π‘₯^2 + 1))/((π‘₯^2 + 1)(π‘₯ βˆ’1)) Canceling denominator 1 = (𝐴π‘₯ + 𝐡)(π‘₯ βˆ’ 1) + 𝐢 (π‘₯^2 + 1) Putting x = 1 1 = (𝐴(1) + 𝐡)(1βˆ’1) + 𝐢 ((βˆ’1)^2 + 1) 1 = (𝐴+𝐡)(0)+ 𝐢 (1+1) 1 = 2𝐢 𝐢=1/2 Putting x = 0 1 = (𝐴π‘₯ + 𝐡)(π‘₯ βˆ’ 1) + 𝐢 (π‘₯^2 + 1) 1 = (𝐴(0) + 𝐡)(0βˆ’1) + 𝐢 (0^2+1) 1 = (𝐡)(βˆ’1) + 𝐢 (1) 1 = 𝐢 βˆ’"B" B =πΆβˆ’1 B =1/2 βˆ’1 B =(βˆ’1)/2 Putting x = βˆ’ 1 1 = (𝐴π‘₯ + 𝐡)(π‘₯ βˆ’ 1) + 𝐢 (π‘₯^2 + 1) 1 = (𝐴(βˆ’1)+ 𝐡)(βˆ’1βˆ’1) + 𝐢 ((βˆ’1)^2+1) 1 = (βˆ’π΄+𝐡)(βˆ’2)+𝐢 (1+1) 1 = (π΄βˆ’π΅)2+𝐢 (2) 1/2=π΄βˆ’π΅+𝐢 𝐴=1/2+π΅βˆ’πΆ 𝐴 =1/2βˆ’1/2βˆ’1/2 𝐴 =(βˆ’1)/2 Hence we can write 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) )= (𝐴π‘₯ + 𝐡)/(π‘₯^2 + 1) + 𝐢/(π‘₯ βˆ’ 1) 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) ) = (βˆ’ 1/2 π‘₯ βˆ’ 1/2)/(π‘₯^2 + 1) + (1/2)/(π‘₯ βˆ’ 1) = (βˆ’1)/2 ( π‘₯)/(π‘₯^2 + 1) βˆ’1/2 1/(π‘₯^2 + 1)+ 1/2(π‘₯ βˆ’ 1) Hence we can write 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) )= (𝐴π‘₯ + 𝐡)/(π‘₯^2 + 1) + 𝐢/(π‘₯ βˆ’ 1) 1/((π‘₯^2 + 1) (π‘₯ βˆ’ 1) ) = (βˆ’ 1/2 π‘₯ βˆ’ 1/2)/(π‘₯^2 + 1) + (1/2)/(π‘₯ βˆ’ 1) = (βˆ’1)/2 ( π‘₯)/(π‘₯^2 + 1) βˆ’1/2 1/(π‘₯^2 + 1)+ 1/2(π‘₯ βˆ’ 1) Therefore, we can write I=∫1β–’γ€–(π‘₯+1)+1/(π‘₯^2 + 1)(π‘₯ βˆ’ 1) 𝑑π‘₯γ€— =∫1β–’[(π‘₯+1)βˆ’1/2 π‘₯/((π‘₯^2 + 1) ) 𝑑π‘₯βˆ’βˆ«1β–’γ€–1/2 1/(π‘₯^2 + 1) 𝑑π‘₯+∫1β–’γ€–1/2 1/((π‘₯ βˆ’ 1) ) 𝑑π‘₯γ€—γ€—] =π‘₯^2/2+π‘₯βˆ’1/2 ∫1β–’γ€–π‘₯/(π‘₯^2 + 1)βˆ’1/2 ∫1β–’γ€–1/(π‘₯^2 + 1) 𝑑π‘₯+1/2 ∫1β–’γ€–1/(π‘₯ βˆ’ 1) 𝑑π‘₯γ€—γ€—γ€— ∴ I = π‘₯^2/2+π‘₯ – 1/2 I"1 βˆ’ " 1/2 I"2 + " 1/2 I"3" Solving π‘°πŸ I1=∫1β–’γ€–π‘₯/(π‘₯^2 + 1) 𝑑π‘₯γ€— Put 𝑑=π‘₯^2+1 Differentiating w.r.t. π‘₯ 𝑑𝑑/𝑑π‘₯=2π‘₯+0 𝑑𝑑/2π‘₯=𝑑π‘₯ Therefore, ∫1β–’γ€–(π‘₯ 𝑑π‘₯)/(π‘₯^2 + 1)=∫1β–’π‘₯/𝑑 𝑑𝑑/2π‘₯γ€—=∫1β–’1/2 𝑑𝑑/𝑑=1/2 π‘™π‘œπ‘”|𝑑|+𝐢1 Putting 𝑑=π‘₯^2+1 =1/2 π‘™π‘œπ‘”|π‘₯^2+1|+𝐢1 And, I2=∫1β–’γ€–1/(π‘₯^2 + 1) 𝑑π‘₯γ€—=tan^(βˆ’1)⁑〖π‘₯+γ€— 𝐢2 I3=∫1β–’γ€–1/(π‘₯ βˆ’1) 𝑑π‘₯γ€—=π‘™π‘œπ‘”|π‘₯βˆ’1|+𝐢3 Hence 𝐼=π‘₯^2/2+π‘₯βˆ’1/2 𝐼1βˆ’1/2 𝐼2+1/2 𝐼3 =π‘₯^2/2+π‘₯βˆ’1/2 (1/2 π‘™π‘œπ‘”|π‘₯^2+1|+𝐢1)βˆ’1/2 (γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1) (π‘₯)+C_2 )βˆ’1/2 (π‘™π‘œπ‘”|π‘₯βˆ’1|+𝐢3) =π‘₯^2/2+π‘₯βˆ’1/4 π‘™π‘œπ‘”|π‘₯^2+1|+𝐢1/2βˆ’1/2 tan^(βˆ’1)⁑〖π‘₯ 𝐢2/2+1/2 π‘™π‘œπ‘”|π‘₯βˆ’1|+𝐢3/2γ€— =𝒙^𝟐/𝟐+𝒙+𝟏/𝟐 π’π’π’ˆ|π’™βˆ’πŸ|βˆ’πŸ/πŸ’ π’π’π’ˆ(𝒙^𝟐+𝟏)βˆ’πŸ/𝟐 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑〖𝒙+π‘ͺγ€—

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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.